CODEWITHSUNDEEP
phpmysql
John
John

Reputation: 4944

Images Not Being Inserted into MySQL database

I am trying to use the form below to insert images into a MySQL database. However, nothing is being put into the MySQL table. Any idea why it's not working?

Thanks in advance,

John

EDIT: The image is $image1.

The form:

echo '<form enctype="multipart/form-data" action="http://www...com/.../submit2.php" method="post"> 
    <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid">  

    <div class="submissiontitle"><label for="title">Blog Post Title:</label></div> 
    <div class="submissionfield"><input class="checkMax3" name="title" type="title" id="title" maxlength="80"></div>         

    <div class="texttitle"><label for="text1">Blog Post Text 1:</label></div> 
    <div class="textfield"><textarea class="checkMax" name="text1" type="comment" id="text1" maxlength="10000"></textarea></div>

    <div class="imagetitle"><label for="image1">Image 1:</label></div> 
    <div class="imagefield"><input type="file" name="image1" /></div>   

    <div class="hyperlinktitle"><label for="url1">Hyperlink 1:</label></div> 
    <div class="hyperlinkfield "><input name="url1" type="title" id="url1" maxlength="200"></div>       

    <div class="submissionbutton"><input name="submit" type="submit" value="Submit"></div> 
</form>
';

The query on submit2.php:

$remove_array = array('http://www.', 'http://', 'https://', 'https://www.', 'www.');    
$uid = $_POST['uid'];    
$title = $_POST['title'];    
$text1 = $_POST['text1'];    
$image1 = $_POST['image1'];    
$image1 = $_FILES['image1'];    
$url1 = $_POST['url1'];    

        $submissionid =  mysql_insert_id();       

$info1 = getImageSize($image1['tmp_name']);    
$queryimage1 = sprintf(
            "insert into images (NULL, submissionid, filename, mime_type, file_size, file_data)
                values ('%s', '%s', '%s', %d, '%s')",
            $submissionid,
            mysql_real_escape_string($image1['name']),
            mysql_real_escape_string($info1['mime']),
            $image1['size'],
            mysql_real_escape_string(
                file_get_contents($image1['tmp_name'])
            )
        )or die(mysql_error());    

mysql_query($queryimage1);

Upvotes: 1

Views: 274

Answers (3)

Hacker
Hacker

Reputation: 7906

i feel you should use

$image1 = $_FILES['image1'];

$newImage = file_get_contents($image1);

and use $newImage in your Mysql query

sorry it was not complete

update

$image1 = $_FILES['image1']['tmp_name'];

$newImage = file_get_contents($image1);

Upvotes: 0

blankabout
blankabout

Reputation: 2637

Depending on how large the images are, you may to have change MySQL's max_allowed_packet value. The default is 1MB, any larger images will result in an error and no data stored.

Upvotes: 0

Marc B
Marc B

Reputation: 360892

Did you check if the file upload succeeded?

if ($_FILES['image1']['error'] !== UPLOAD_ERR_OK) {
   die("File didn't upload, error code is {$_FILES['image1']['error']");
}

Did you check if the query succeeded?

$result = mysql_query($queryimage1);
if ($result === FALSE) {
   die("Query failed: " . mysql_error());
}

Honestly... why are there so many posts here without even the most basic simplest of error handling/checking? It's as if people expect code to magically work regardless of any errors, so there's no point in putting in error handling.

Upvotes: 2

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