Question

i try to insert multiple row in my database, that image not save to folder and
i get this error

Blockquote ArrayArrayYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘Chrysanthemum.jpg’), (‘Hydrangeas.jpg’)’, ‘(‘Chrysanthemum.jpg’), (‘Hydrangeas.j’ at line 1

this is my code :

$hopid = $_POST[photo_hop_id];
$title = $_POST['photo_name_id'];

if(!is_array($title)) {
    $title = array();
}

$titleds = "('" . implode("'), ('", $title) . "')";

$tmp_file = $_FILES['ne_photo_image']['tmp_name'];
$file = $_FILES['ne_photo_image']['name'];  

if(!is_array($tmp_file)) {
    $tmp_file = array();
}
if(!is_array($file)) {
    $file = array();
}

$sementara = "('" . implode("'), ('", $tmp_file) . "')";
$filed = "('" . implode("'), ('", $file) . "')";    

$tmp_file1 = $_FILES['fe_photo_image']['tmp_name'];
$file1 = $_FILES['fe_photo_image']['name']; 

if(!is_array($tmp_file1)) {
    $tmp_file = array();
}
if(!is_array($file1)) {
    $file = array();
}

$sementara1 = "('" . implode("'), ('", $tmp_file) . "')";
$filed1 = "('" . implode("'), ('", $file) . "')";   

if(!move_uploaded_file($sementara, 'image/' . $filed)) {
    echo $_FILES["ne_photo_image"]["error"];
}

if(!move_uploaded_file($sementara1, 'image/' . $filed1)) {
    echo $_FILES["fe_photo_image"]["error"];
}

$y = "INSERT INTO photo VALUES (null, '".$filed."', '".$filed1."', '".$hopid."', '".$titleds."')";
$z = mysql_query($y) or die (mysql_error());
if($z) {
    $msg = "Data sudah ditambahkan";
}
else {
    $msg = "Data tidak bisa dimasukkan";
}

echo print_r($y);

and this is my form:

<form method="post" enctype="multipart/form-data">
    <table border="0"cellpadding="0" cellspacing="0" width= "100%">
        <tr>
            <td>Hop Name :<?echo "$data[hop_name]"?>
                <input type='hidden' name='photo_hop_id' value='<?echo"$data[hop_id]"?>'>
            </td>
        </tr>
        <table border="0"cellpadding="0" cellspacing="0" width= "100%">
            <tr>
                <td cellpadding="0" cellspacing="0" width= "50%"> 
                    Near End Site Name : <?echo "$data[ne_site_name]" ?>
                    </br>
                    Near End Site Id : <?echo "$data[ne_site_code]" ?>
                </td>
                <td cellpadding="0" cellspacing="0" width= "50%"> 
                    Far End Site Name : <?echo "$data[fe_site_name]" ?>
                    </br>
                    Far End Site Id : <?echo "$data[fe_site_code]"?>
                </td>
            </tr>   
            <tr>
                <td cellpadding="0" cellspacing="0" width= "50%"> 
                    <?  $pm1= mysql_query("SELECT photo_name FROM photo_name WHERE photo_name_id = 1");
                    $dpm1 = mysql_fetch_array ($pm1);echo"$dpm1[0]" ?> 
                    <input type='hidden' name='photo_name_id[]' value='<?echo"$dpm1[0]"?>'> :  
                    <input type="file" name="ne_photo_image[]">
                </td>
                <td cellpadding="0" cellspacing="0" width= "50%"> 
                    <?echo "$dpm1[0]"?> : <input type="file" name="fe_photo_image[]">
                </td>
            </tr>   
            <tr>
                <td cellpadding="0" cellspacing="0" width= "50%"> 
                    <? $pm1= mysql_query("SELECT photo_name FROM photo_name WHERE photo_name_id = 2");
                    $dpm1 = mysql_fetch_array ($pm1);echo"$dpm1[0]" ?> 
                    <input type='hidden' name='photo_name_id[]' value='<?echo"$dpm1[0]"?>'> :  
                    <input type="file" name="ne_photo_image[]">
                </td>
                <td cellpadding="0" cellspacing="0" width= "50%"> 
                    <?echo "$dpm1[0]"?> : <input type="file" name="fe_photo_image[]">
                </td>
            </tr>   
        </table>
    </table>

    <input type="submit" value="insert" />
</form>

many thanks for help

Answer 1

Your error comes clearly from this line :

$y = "INSERT INTO photo VALUES (null, '".$filed."', '".$filed1."', '".$hopid."', '".$titleds."')";

For example, your var $filed is an imploded array, which looks like ('foo'), ('bar'). Finally, your request will looks like INSERT INTO photo VALUE (null, '('foo'), ('bar')', [...]);.

You must escape simple quotes in $filed, $filed1, $hopid, $titleds, but I’m pretty sure your request is totally wrong.

Can you give us the photo table schema structure please ?

EDIT :

structure given : (photo_id,ne_photo_image,fe_photo_image,hop_id,title)

To insert multiple rows in a table, you must use this kind of syntax :

INSERT INTO photo(`photo_id`, `ne_photo_image`, `fe_photo_image`, `hop_id`, `title`)
VALUES
    (null, 'foo', 'bar', 'baz', 'qux'),
    (null, 'foo', 'bar', 'baz', 'qux'),
    (null, 'foo', 'bar', 'baz', 'qux');

Aka, you must insert your datas row by row, not column by column (what you’re trying to attempt right now).

Your code should looks like this (this is definitely not a full piece of code, I’m not sure to understand all your code, then I give you some track) :

<?php 
   $hopid = $_POST['photo_hop_id'];

   $titles = $_POST['photo_name_id'];
   $ne_photo_images = $_FILES['ne_photo_image']['tmp_name'];
   $fe_photo_images = $_FILES['fe_photo_image']['tmp_name'];

   /*
   I assume $titles, $ne_photo_images and $fe_photo_images 
   have the same number of elements, in the right order.
   */

   $sql = "INSERT INTO photo(`photo_id`, `ne_photo_image`, `fe_photo_image`, `hop_id`,    `title`) VALUES";
   for($i = 0, $l = sizeof($titles) ; $i < $l ; $i++)
   {
      //adding row datas
      $sql .= " (null, 
                 '".$ne_photo_images [$i]."', 
                 '".$fe_photo_images [$i]."', 
                 '".$hopid."', 
                 '".$titles[$i]."')";
      if($i < $l - 1)
         $sql .= ",";
   }

   if(mysql_query($sql))
      //get happy
   else
      //take a coffee
?>

Again, this code is far from perfect, I’m just trying to explain you how to approach the problem.