Check if image has been selected for upload
0
I have a form were a user can enter text and also upload images.
Saving the values and images works fine. If the user does not select an image, I get the error message that no file has been selected (Error 4).
How can I run the upload image part only if the user selected an image?
I have tried:
if (!empty($_FILES['files']['name'])){ Upload the Image }
if (!empty($_FILES['files'])){ Upload the Image }
if (!empty($_FILES['files']['name'])) { Upload the Image }
if ($_FILES['files']['error'] == UPLOAD_ERR_OK) { Upload the Image }
This the form:
<input type="file" name="files" title="Title" maxlength="10" accept="gif|jpg|jpeg|png|tiff">
php image-uploading
asked Nov 12 at 19:38
Beji
277
add a comment |
0
I have a form were a user can enter text and also upload images.
Saving the values and images works fine. If the user does not select an image, I get the error message that no file has been selected (Error 4).
How can I run the upload image part only if the user selected an image?
I have tried:
if (!empty($_FILES['files']['name'])){ Upload the Image }
if (!empty($_FILES['files'])){ Upload the Image }
if (!empty($_FILES['files']['name'])) { Upload the Image }
if ($_FILES['files']['error'] == UPLOAD_ERR_OK) { Upload the Image }
This the form:
<input type="file" name="files" title="Title" maxlength="10" accept="gif|jpg|jpeg|png|tiff">
php image-uploading
asked Nov 12 at 19:38
Beji
277
Possible duplicate of $_FILES field 'tmp_name' has no value on .JPG file extension
– Martin Zeitler
Nov 12 at 19:45
add a comment |
0
0
0
I have a form were a user can enter text and also upload images.
Saving the values and images works fine. If the user does not select an image, I get the error message that no file has been selected (Error 4).
How can I run the upload image part only if the user selected an image?
I have tried:
if (!empty($_FILES['files']['name'])){ Upload the Image }
if (!empty($_FILES['files'])){ Upload the Image }
if (!empty($_FILES['files']['name'])) { Upload the Image }
if ($_FILES['files']['error'] == UPLOAD_ERR_OK) { Upload the Image }
This the form:
<input type="file" name="files" title="Title" maxlength="10" accept="gif|jpg|jpeg|png|tiff">
php image-uploading
asked Nov 12 at 19:38
Beji
277
I have a form were a user can enter text and also upload images.
Saving the values and images works fine. If the user does not select an image, I get the error message that no file has been selected (Error 4).
How can I run the upload image part only if the user selected an image?
I have tried:
if (!empty($_FILES['files']['name'])){ Upload the Image }
if (!empty($_FILES['files'])){ Upload the Image }
if (!empty($_FILES['files']['name'])) { Upload the Image }
if ($_FILES['files']['error'] == UPLOAD_ERR_OK) { Upload the Image }
This the form:
<input type="file" name="files" title="Title" maxlength="10" accept="gif|jpg|jpeg|png|tiff">
php image-uploading
php image-uploading
asked Nov 12 at 19:38
Beji
277
asked Nov 12 at 19:38
Beji
277
asked Nov 12 at 19:38
Beji
277
asked Nov 12 at 19:38
Beji
277
asked Nov 12 at 19:38
Beji
277
277
Possible duplicate of $_FILES field 'tmp_name' has no value on .JPG file extension
– Martin Zeitler
Nov 12 at 19:45
add a comment |
Possible duplicate of $_FILES field 'tmp_name' has no value on .JPG file extension
– Martin Zeitler
Nov 12 at 19:45
Possible duplicate of $_FILES field 'tmp_name' has no value on .JPG file extension
– Martin Zeitler
Nov 12 at 19:45
Possible duplicate of $_FILES field 'tmp_name' has no value on .JPG file extension
– Martin Zeitler
Nov 12 at 19:45
add a comment |
1 Answer
1
active
oldest
votes
1
Use is_uploaded_file() function to check if the user has uploaded any file or not, and then process inputs accordingly, like this:
if(is_uploaded_file($_FILES['files']['tmp_name'][0])){
// user has uploaded a file
}else{
// user hasn't uploaded anything
}
Above solution code is based on your name attribute of input tag,
<input ... name="files" ... />
If it was <input ... name="files" ... /> then the if condition would be like this:
if(is_uploaded_file($_FILES['files']['tmp_name'])){
...
}else{
...
}
Sidenote: Use var_dump($_FILES); to see the complete array structure.
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
If I use if(is_uploaded_file($_FILES['files']['tmp_name'])){...} the code got not called even if I select an image for upload.
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) without selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(0) "" } ["type"]=> array(1) { [0]=> string(0) "" } ["tmp_name"]=> array(1) { [0]=> string(0) "" } ["error"]=> array(1) { [0]=> int(4) } ["size"]=> array(1) { [0]=> int(0) } } }
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) with selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(63) "media.media.56852c99-74ee-4e6e-bf29-b37b0ce6bb46.normalized.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(19) "/home/tmp/phpvyOI7C" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(27933) } } }
– Beji
Nov 12 at 20:12
@Beji, as I said in my updated answer; given the current<input ..>tag, yourifblock should be like this:if(is_uploaded_file($_FILES['files']['tmp_name'][0])){ ...
– Rajdeep Paul
Nov 12 at 20:15
thanks, I forgot to add [0]
– Beji
Nov 12 at 20:16
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Use is_uploaded_file() function to check if the user has uploaded any file or not, and then process inputs accordingly, like this:
if(is_uploaded_file($_FILES['files']['tmp_name'][0])){
// user has uploaded a file
}else{
// user hasn't uploaded anything
}
Above solution code is based on your name attribute of input tag,
<input ... name="files" ... />
If it was <input ... name="files" ... /> then the if condition would be like this:
if(is_uploaded_file($_FILES['files']['tmp_name'])){
...
}else{
...
}
Sidenote: Use var_dump($_FILES); to see the complete array structure.
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
If I use if(is_uploaded_file($_FILES['files']['tmp_name'])){...} the code got not called even if I select an image for upload.
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) without selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(0) "" } ["type"]=> array(1) { [0]=> string(0) "" } ["tmp_name"]=> array(1) { [0]=> string(0) "" } ["error"]=> array(1) { [0]=> int(4) } ["size"]=> array(1) { [0]=> int(0) } } }
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) with selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(63) "media.media.56852c99-74ee-4e6e-bf29-b37b0ce6bb46.normalized.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(19) "/home/tmp/phpvyOI7C" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(27933) } } }
– Beji
Nov 12 at 20:12
@Beji, as I said in my updated answer; given the current<input ..>tag, yourifblock should be like this:if(is_uploaded_file($_FILES['files']['tmp_name'][0])){ ...
– Rajdeep Paul
Nov 12 at 20:15
thanks, I forgot to add [0]
– Beji
Nov 12 at 20:16
add a comment |
1
Use is_uploaded_file() function to check if the user has uploaded any file or not, and then process inputs accordingly, like this:
if(is_uploaded_file($_FILES['files']['tmp_name'][0])){
// user has uploaded a file
}else{
// user hasn't uploaded anything
}
Above solution code is based on your name attribute of input tag,
<input ... name="files" ... />
If it was <input ... name="files" ... /> then the if condition would be like this:
if(is_uploaded_file($_FILES['files']['tmp_name'])){
...
}else{
...
}
Sidenote: Use var_dump($_FILES); to see the complete array structure.
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
If I use if(is_uploaded_file($_FILES['files']['tmp_name'])){...} the code got not called even if I select an image for upload.
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) without selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(0) "" } ["type"]=> array(1) { [0]=> string(0) "" } ["tmp_name"]=> array(1) { [0]=> string(0) "" } ["error"]=> array(1) { [0]=> int(4) } ["size"]=> array(1) { [0]=> int(0) } } }
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) with selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(63) "media.media.56852c99-74ee-4e6e-bf29-b37b0ce6bb46.normalized.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(19) "/home/tmp/phpvyOI7C" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(27933) } } }
– Beji
Nov 12 at 20:12
@Beji, as I said in my updated answer; given the current<input ..>tag, yourifblock should be like this:if(is_uploaded_file($_FILES['files']['tmp_name'][0])){ ...
– Rajdeep Paul
Nov 12 at 20:15
thanks, I forgot to add [0]
– Beji
Nov 12 at 20:16
add a comment |
1
1
1
Use is_uploaded_file() function to check if the user has uploaded any file or not, and then process inputs accordingly, like this:
if(is_uploaded_file($_FILES['files']['tmp_name'][0])){
// user has uploaded a file
}else{
// user hasn't uploaded anything
}
Above solution code is based on your name attribute of input tag,
<input ... name="files" ... />
If it was <input ... name="files" ... /> then the if condition would be like this:
if(is_uploaded_file($_FILES['files']['tmp_name'])){
...
}else{
...
}
Sidenote: Use var_dump($_FILES); to see the complete array structure.
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
Use is_uploaded_file() function to check if the user has uploaded any file or not, and then process inputs accordingly, like this:
if(is_uploaded_file($_FILES['files']['tmp_name'][0])){
// user has uploaded a file
}else{
// user hasn't uploaded anything
}
Above solution code is based on your name attribute of input tag,
<input ... name="files" ... />
If it was <input ... name="files" ... /> then the if condition would be like this:
if(is_uploaded_file($_FILES['files']['tmp_name'])){
...
}else{
...
}
Sidenote: Use var_dump($_FILES); to see the complete array structure.
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
edited Nov 12 at 20:01
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
answered Nov 12 at 19:47
Rajdeep Paul
15.9k31229
15.9k31229
If I use if(is_uploaded_file($_FILES['files']['tmp_name'])){...} the code got not called even if I select an image for upload.
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) without selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(0) "" } ["type"]=> array(1) { [0]=> string(0) "" } ["tmp_name"]=> array(1) { [0]=> string(0) "" } ["error"]=> array(1) { [0]=> int(4) } ["size"]=> array(1) { [0]=> int(0) } } }
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) with selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(63) "media.media.56852c99-74ee-4e6e-bf29-b37b0ce6bb46.normalized.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(19) "/home/tmp/phpvyOI7C" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(27933) } } }
– Beji
Nov 12 at 20:12
@Beji, as I said in my updated answer; given the current<input ..>tag, yourifblock should be like this:if(is_uploaded_file($_FILES['files']['tmp_name'][0])){ ...
– Rajdeep Paul
Nov 12 at 20:15
thanks, I forgot to add [0]
– Beji
Nov 12 at 20:16
add a comment |
If I use if(is_uploaded_file($_FILES['files']['tmp_name'])){...} the code got not called even if I select an image for upload.
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) without selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(0) "" } ["type"]=> array(1) { [0]=> string(0) "" } ["tmp_name"]=> array(1) { [0]=> string(0) "" } ["error"]=> array(1) { [0]=> int(4) } ["size"]=> array(1) { [0]=> int(0) } } }
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) with selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(63) "media.media.56852c99-74ee-4e6e-bf29-b37b0ce6bb46.normalized.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(19) "/home/tmp/phpvyOI7C" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(27933) } } }
– Beji
Nov 12 at 20:12
@Beji, as I said in my updated answer; given the current<input ..>tag, yourifblock should be like this:if(is_uploaded_file($_FILES['files']['tmp_name'][0])){ ...
– Rajdeep Paul
Nov 12 at 20:15
thanks, I forgot to add [0]
– Beji
Nov 12 at 20:16
If I use if(is_uploaded_file($_FILES['files']['tmp_name'])){...} the code got not called even if I select an image for upload.
– Beji
Nov 12 at 20:11
If I use if(is_uploaded_file($_FILES['files']['tmp_name'])){...} the code got not called even if I select an image for upload.
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) without selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(0) "" } ["type"]=> array(1) { [0]=> string(0) "" } ["tmp_name"]=> array(1) { [0]=> string(0) "" } ["error"]=> array(1) { [0]=> int(4) } ["size"]=> array(1) { [0]=> int(0) } } }
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) without selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(0) "" } ["type"]=> array(1) { [0]=> string(0) "" } ["tmp_name"]=> array(1) { [0]=> string(0) "" } ["error"]=> array(1) { [0]=> int(4) } ["size"]=> array(1) { [0]=> int(0) } } }
– Beji
Nov 12 at 20:11
If I run var_dump($_FILES) with selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(63) "media.media.56852c99-74ee-4e6e-bf29-b37b0ce6bb46.normalized.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(19) "/home/tmp/phpvyOI7C" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(27933) } } }
– Beji
Nov 12 at 20:12
If I run var_dump($_FILES) with selecting a file: array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(63) "media.media.56852c99-74ee-4e6e-bf29-b37b0ce6bb46.normalized.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(19) "/home/tmp/phpvyOI7C" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(27933) } } }
– Beji
Nov 12 at 20:12
@Beji, as I said in my updated answer; given the current
<input ..> tag, your if block should be like this: if(is_uploaded_file($_FILES['files']['tmp_name'][0])){ ...– Rajdeep Paul
Nov 12 at 20:15
@Beji, as I said in my updated answer; given the current
<input ..> tag, your if block should be like this: if(is_uploaded_file($_FILES['files']['tmp_name'][0])){ ...– Rajdeep Paul
Nov 12 at 20:15
thanks, I forgot to add [0]
– Beji
Nov 12 at 20:16
thanks, I forgot to add [0]
– Beji
Nov 12 at 20:16
add a comment |
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Possible duplicate of $_FILES field 'tmp_name' has no value on .JPG file extension
– Martin Zeitler
Nov 12 at 19:45