PHP Multiple file upload with multiple inputs - php

Question

I'm new to PHP. Now i have a problem with files upload.
All files are moved, but It didn't store file's name to database.
and it didn't show error. I have no idea to fix this one. Please help me out.
<form method="post" action="index.php?insert_ads" enctype="multipart/form-data">
<input type="file" name="b1" id="b1"/>
<b>Link</b></br>
<input type="text" id="b1l" name="b1l" class="form-control"/></br>
<b>Home Small</b> <b style="color: blue;">100 x 100 px</b></br>
<input type="userfile" name="b2" id="b2"/><br>
<b>Link</b></br>
<input type="text" id="b2l" name="b2l" class="form-control"/></br>
<input type="submit" name="submit" value="Publish"/>
</form></br>
<?php
if(isset($_POST['submit'])){
$b1 = $_FILES['b1']['name'];
$tmp1 = $_FILES['b1']['tmp_name'];
$b1l = $_POST['b1l'];
$b2 = $_FILES['b2']['name'];
$tmp2 = $_FILES['b2']['tmp_name'];
$b2l = $_POST['b2l'];
move_uploaded_file($tmp1,"ads/$b1");
move_uploaded_file($tmp2,"ads/$b2");
$insert_posts = "insert into ads (b1,b2) value ('$b1','$b2')";
$run_posts = mysql_query($insert_posts);
}
?>

Answer 1

Notwithstanding any issues about using mysql_query or injection attacks, there are a number of things that could be going wrong here.
One option is that the query is executing, but you haven't assigned the $b1 and $b2 variables correctly. This would be the case if rows are being added to the database, but the rows are empty (e.g., SELECT b1, b2 FROM db.ads" returns rows of '',''); in that case, you probably just aren't extracting the name attribute from the $_FILES variable correctly. You can run var_dump($_FILES); to see more information about it and figure out what you need to get.
Another possibility is that the query is not executing. Again, this may be for a couple of reasons -- maybe (somehow) it's not reaching that point in the code. You can test that like so:
$insert_posts = "insert into ads (b1,b2) value ('$b1','$b2')";
echo $insert_posts; // if this shows up, you're running the next line also
$run_posts = mysql_query($insert_posts);
Another option is that your error reporting level is not capturing an error. A likely cause of this is that you have not connected to the database -- according to the mysql_query documentation...
If no connection is found or established, an E_WARNING level error is generated.
A E_WARNING level error will allow the program to continue to execute unless you have configured your program to behave differently.
Finally, you may have a syntax error (and indeed it seems you do -- VALUES, not VALUE); according to the documentation, mysql_query returns false on error -- it does not throw an error.
You can rig it to do so by testing for false and using the mysql_error function to get the error:
$run_posts = mysql_query($insert_posts);
if ($run_posts === false) {
trigger_error("Error in SQL!\n" + mysql_error(), E_USER_ERROR);
}