我需要从php数据中提取URL,我该如何实现?从ajax响应中提取数据
PHP
$query = 'SELECT * FROM picture LIMIT 3';
$result = mysql_query($query);
while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) {
$url.=$rec['pic_location'].";";
}
echo json_encode($url);
阿贾克斯
<script type="text/javascript">
$(document).ready(function() {
$(".goButton").click(function() {
var dir = $(this).attr("id");
var imId = $(".theImage").attr("id");
$.ajax({
url: "viewnew.php",
data: {
current_image: imId,
direction : dir
},
success: function(ret){
console.log(ret);
var arr = ret;
alert("first: " + arr[0] + ", second: " + arr[1]);
alert(arr[0]);
$(".theImage").attr("src", +arr[0]);
if ('prev' == dir) {
imId ++;
} else {
imId --;
}
$("#theImage").attr("id", imId);
}
});
});
});
</script>
警报消息不工作它只是打印HT(我觉得这些都是HTTP:// ...)
什么的的console.log节目吗? –
@FlorianMargaine - 无法加载资源:服务器以$(“。theImage”)。attr(“src”,+ arr [0]);“'和' alert(arr [0]);'显示正确的URL – Yahoo