一切都在执行，但查询...我错过了什么？

Question

This is a file that receives uploaded images from another page. The images are received and saved on the server (that part is working) the query is not. I manually typed the query into phpMyAdmin with dummy values for the file name and it works. I put that dummy query into this code and it doesn't work. I can't figure out what I'm doing wrong.

<?php
$ds  = DIRECTORY_SEPARATOR;  
$storeFolder = 'img';   
if (!empty($_FILES)) {
    $tempFile = $_FILES['file']['tmp_name'];              
    $targetPath = $_SERVER['DOCUMENT_ROOT']. $ds. $storeFolder . $ds;  
    $targetFile =  $targetPath. $_FILES['file']['name'];  
    move_uploaded_file($tempFile,$targetFile); 
    mysql_query("INSERT INTO `photos` (photo_id, file_name) VALUES ('', '".$_FILES['file']['name']."');"); 
}
?>  

Answer 1

Assuming that photo_id is auto incremented here are some fixes to your code:

<?php
$ds  = DIRECTORY_SEPARATOR;  
$storeFolder = 'img';   
if (!empty($_FILES)) {
    $name = $_FILES['file']['name'];
    $tempFile = $_FILES['file']['tmp_name'];              
    $targetPath = $_SERVER['DOCUMENT_ROOT']. $ds. $storeFolder . $ds;  
    $targetFile =  $targetPath. $_FILES['file']['name'];  
    move_uploaded_file($tempFile,$targetFile); 
    $query = "INSERT INTO photos (photo_id, file_name) VALUES (NULL, '".$name."')"; 
    $result = mysql_query($query);
    if (!$result) die ("Database access failed: " . mysql_error());
}
?>  

This way you should be able to debug your code. Now if you need to get the result from a Javascript call, you should use AJAX.

Check here:

Send a request to a php page and then get back results using ajax