Question

我需要从php数据中提取URL，我该如何实现？

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);

Ajax

 <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>

警报消息无法正常打印H T（我认为这些是http：// ...）

Answer 1

您将返回一个未解析为JSON的字符串。只需将dataType: "json"添加到ajax设置即可。

由于你在javascript中将它作为数组读取，你应该像这样返回它：

while ($rec = mysql_fetch_array($result,  MYSQL_ASSOC)) {
    $url[] = $rec['pic_location'];
}

Answer 2

您正在PHP中发送一个字符串，并希望在javascript中将数组作为响应。将PHP改为

while ($rec = mysql_fetch_array($result,  MYSQL_ASSOC)) {
    $url[] = $rec['pic_location'];
}

和javascript到

$.ajax({
      url: "viewnew.php",
      dataType: "JSON",
      data: {
         current_image: imId,
         direction    : dir
      },
      success: function(ret){
          console.log(ret[0]);
          var arr = ret;
          alert(arr);  
          alert("first: " + arr[0] + ", second: " + arr[1]);    // THIS IS NOT WORKING!!!!
          if ('prev' == dir) {
            imId ++;
         } else {
            imId --;
         }
         $("#theImage").attr("id", imId);
      }
   });

从ajax响应中提取数据

2 个答案: