将数据从php传递回ajax

如何将数据从然后行的PHP传递回ajax？

PHP

$query = 'SELECT * FROM picture order by rand() LIMIT 10'; $result = mysql_query($query); while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) { $url[]=$rec['pic_location']; $name[]=$rec['name']; $age[]=$rec['age']; $gender[]=$rec['gender']; } echo json_encode($url); echo json_encode($name); echo json_encode($age); echo json_encode($gender); 

阿贾克斯

 $(".goButton").click(function() { var dir = $(this).attr("id"); var imId = $(".theImage").attr("id"); $.ajax({ url: "viewnew.php", dataType: "json", data: { current_image: imId, direction : dir }, success: function(ret){ console.log(ret); var arr = ret; alert("first image url: " + arr[0][0] + ", second image url: " + arr[0][1]); // This code isnt working alert("first image Name: " + arr[1][0] + ", second image name: " + arr[1][1]); $(".theImage").attr("src", arr[0]); if ('prev' == dir) { imId ++; } else { imId --; } $("#theImage").attr("id", imId); } }); }); });  

我的问题是如何在这里显示值？ 警告信息给我“未定义”？

Answer 1

你可以沿着这些方向做点什么。

PHP

 $query = 'SELECT * FROM picture order by rand() LIMIT 10'; $res = mysql_query($query); $pictures = array(); while ($row = mysql_fetch_array($res)) { $picture = array( "pic_location" => $row['pic_location'], "name" => $row['name'], "age" => $row['age'], "gender" => $row['gender'] ); $pictures[] = $picture; } echo json_encode($pictures); 

JS

 ... $.ajax({ ... dataType: "json", ... success: function(pictures){ $.each(pictures, function(idx, picture){ // picture.pic_location // picture.name // picture.age // picture.gender }); } }); ... 

Answer 2

您不能为AJAX响应放置多个echo语句：

 echo json_encode($url); echo json_encode($name); echo json_encode($age); echo json_encode($gender); 

加入您的arrays并发送一个响应：

 $arr = $url + $name + $age + $gender; echo json_encode($arr); 

Answer 3

您可以使用单个数组轻松完成此操作：

 $pics = array(); while ($rec = mysql_fetch_array($result, MYSQL_ASSOC)) { $pics[$rec['id']]['url'] = $rec['pic_location']; $pics[$rec['id']]['name']=$rec['name']; $pics[$rec['id']]['age']=$rec['age']; $pics[$rec['id']]['gender']=$rec['gender']; } echo json_encode($pics);