خرید بک لینک

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This is my html code. everytime the button is clicked the jquery sript will run. This is part of a form.

    <form class="addproducts" method="POST" action="addproducts.php" enctype="multipart/form-data">.
..
..
..
....
..
<div class="multi-img">             
                <div class="img-group">
                    <!-- multi image form-->
                </div>
                <button id="imgs"><span class="glyphicon glyphicon-plus-sign"></span></button>
            </div><!--multi row-->
            <input type="submit" value="Add Product" />
        </form> 
...
...
...
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
    $("button#imgs").click(function(event){
        event.preventDefault();
        $("div.img-group").append('<div class="form-group">'+
                        '<input type="file" name="group-img" />'+
                    '</div>'+
                    '</br>'+
                    '<div class="form-group">'+
                        '<label>Title</label>'+
                        '<input type="input" name="group-imagetitle" />'+
                    '</div>'+
                    '<div class="form-group">'+
                        '<label>Alt Text</label>'+
                        '<input type="input" name="group-imagealt" />'+
                    '</div>'+
                    '</br>');

        });

});
</script>

this is my php code

for ($i=0; $i<count($_FILES['group-img']['name']); $i++) { 

    $prodgp_imgnme=$_FILES['group-img']['name'][$i];
    $prodgp_img=addslashes(file_get_contents($_FILES['group-img']['tmp_name'][$i]));
    $prodgp_imgtitle=$_POST['group-imagetitle'][$i];
    $prodgp_imgalt=$_POST['group-imagealt'][$i];

    mysqli_query($con,"INSERT INTO nfprodimg(prodi_img,*prodi_prodid,prodi_nme,prodi_title,prodi_alt,prodi_collid)
            VALUES('$prodgp_img','$sku','$prodgp_imgnme','$prodgp_imgtitle', '$prodgp_imgalt','$coll')");


}

my loop doesn't seem to work. i tried to echo the number of $_FILES['group-img']['name']. The result is always 1.

asked 1 min ago

برچسب: نویسنده: استخدام کار تاريخ: شنبه 8 خرداد 1395 ساعت: 0:28

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