sameer kumar sameer kumar - 4 months ago 28
SQL Question

Insert multiple image name in a single row in database

Mates Please Help me to solve this problem.
I ve created one from where multiple Image can Insert to a single row of database now my code is working well but it is not possible to store all image in a single data base row it is storing in multiple rows. Here what i ve done till yet.

<?php
mysql_connect("localhost","root","");
mysql_select_db("test");



$uploads_dir = 'photo/';
foreach ($_FILES["image"]["error"] as $key => $error) {
if ($error == UPLOAD_ERR_OK) {
$tmp_name = $_FILES["image"]["tmp_name"][$key];
$name = $_FILES["image"]["name"][$key];
move_uploaded_file($tmp_name, "$uploads_dir/$name");
$sql=mysql_query("INSERT INTO multiimg SET image='$name'");
}
}


?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
<script>
function addmore(num)
{
if(num==1)
{
document.getElementById('field2').style.display='block';
document.getElementById('ni1').style.display='block';
return false;
}
else if(num==2)
{
document.getElementById('field3').style.display='block';
return false;
}

}
</script>
</head>

<body>
<form enctype="multipart/form-data" name="" action="" method="post">
<div id="field1">Enter One Image :<input type="file" name="image[]" id="img1"/><a href="#" onclick="addmore(1)" id="ni1">addmore...</a></div>
<div id="field2" style="display:none;">Enter Two Image :<input type="file" name="image[]" id="img2"/><a href="#" onclick="addmore(2);">add more...</a></div>
<div id="field3" style="display:none;">Enter Three Image :<input type="file" name="image[]" id="img3"/><a href="#" onclick="addmore(3)" id="ni3">addmore...</a></div>
<div id="field4" style="display:none">Enter Forth Image :<input type="file" name="image[]" id="img4"/><a href="#" onclick="addmore(4)" id="ni4">addmore...</a></div>

<input type="submit" name="submit"/>

</form>

</body>
</html>


enter image description here

Answer

Try below code

$images_name ="";
    foreach ($_FILES["image"]["error"] as $key => $error) {
        if ($error == UPLOAD_ERR_OK) {
            $tmp_name = $_FILES["image"]["tmp_name"][$key];
            $name = $_FILES["image"]["name"][$key];
            move_uploaded_file($tmp_name, "$uploads_dir/$name");
            $images_name =$images_name.",".$name;
        }
    }

    $sql=mysql_query("INSERT INTO multiimg(image) values('".$images_name."')");
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