我正在通过ajax获取产品项目的特定列表,通过将其唯一id传递到服务器。现在每个产品都有自己的属性集,我必须在页面上显示产品图像。当我通过jquery设置值时,只有数组中的最后一个值被打印出来。以下是我的编码文件。
images.php
while($fetch = mysql_fetch_array($result))
{
?>
<div class="col-sm-4">
<div class="thumbnail">
<a class="productitemid" href="productpurchase.php?id=<?php echo $fetch['itemID'];?>"><img class="img-responsive productimage" src="uploadedfiles\<?php echo $fetch['imageURL'];?>" alt="<?php echo $fetch['imageURL'];?>" /></a>
<div class="text-center productitemname" style="font-weight:bold;"><?php echo $fetch['itemName']; ?></div>
<div class="badge col-sm-offset-1 productprice"><?php echo $fetch['price']; ?></div>
<span class="col-md-offset-7"><a class="productitemid btn btn-success" href="productpurchase.php?id=<?php echo $fetch['itemID'];?>">BUY</a></span>
</div>
</div>
<?php
}
js文件
$(document).ready(function(){
$('.menProdCatgry').on('click',function(){
$.ajax({
type: "post",
url: "getselectedproducts.php",
data:{
"prodId" : $('.menProdCatgry').attr('prodCatId')
},
dataType: "json",
success: function(data){
console.log(data);
$.each(data, function(){
var getprodId = this.prodId;
var getimageURL = this.imageURL;
var getprice = this.price;
var getitemName = this.itemName;
var getitemID = this.itemID;
$('.productimage').attr('src','uploadedfiles\/'+getimageURL);
$('.productitemname').text(getitemName);
$('.productprice').text(getprice);
$('.productitemid').attr('href','productpurchase.php?id='+getitemID);
});
},
error: function(data){
console.log(data);
}
});
});
});
https://stackoverflow.com/questions/30092649
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