Question

我正在使用ajax（jquery）查询我的数据库，但我收回的数据证明很难处理。我注意到其他人发布了类似的问题，并确定了PHP代码（特别是如何将数据分配给数组）作为问题。但是我找不到解决方案。我返回的数据的内容（当我将返回的数据转换为字符串时）如下所示：

<?php
$replace = array('{COVER_AMT}','{LIABILITY_AMT}','{TOTAL_AMT}', 'This is total amount : 0');
$with = array('90', '90', '0','');

$myString = 'This is Cover Amt : {COVER_AMT} . This is liablity amount :     {LIABILITY_AMT} . This is total amount : {TOTAL_AMT}';


echo str_replace($replace, $with, $myString); 
?>

我真的希望能够从返回的数据中提取单个元素，例如userLastLogin字段。

我的ajax查询：

<?php

$replace = array('{COVER_AMT}','{LIABILITY_AMT}','{TOTAL_AMT}');
$with = array('0', '90', '0');

$myString = 'This is Cover Amt : {COVER_AMT} . This is liablity amount :     {LIABILITY_AMT} . This is total amount : {TOTAL_AMT}';

$myString = str_replace($replace, $with, $myString);
$myString_array = explode("This is",$myString);
foreach($myString_array as $myString_sliced) { 
$pattern = '/(\.?)(This is )(.*?) 0(\s?)(\.?)/i';
$replacement = '';
if($myString_sliced)
echo preg_replace($pattern, $replacement, "This is ".$myString_sliced);
}
?>

我的PHP下面：

[{"id":"1","userName":"admin","userPassword":"admin","userEmail":"admin@admin.com","userRegistrationIP":"","registrationDateTime":"2015-05-28 21:22:54","userLastLogin":"2015-05-28 21:22:54"}]

Answer 1

您可以像这样（客户端）

访问数组中的各个元素

$.ajax({
  ....
  ....
  success: function (res) {

      alert(res[0].userLastLogin);

      // OR

      alert(res[0]['userLastLogin']);

   }
});

Answer 2

这会帮助你

<强> JS

success: function (res) {
var res = [{"id":"1","userName":"admin","userPassword":"admin","userEmail":"admin@admin.com","userRegistrationIP":"","registrationDateTime":"2015-05-28 21:22:54","userLastLogin":"2015-05-28 21:22:54"}];
$.each(res,function(key,value){
     alert(res[key].id);
     alert(res[key].userName);
     alert(res[key].userPassword);
     alert(res[key].userEmail);
     alert(res[key].userRegistrationIP);
     alert(res[key].registrationDateTime);
     alert(res[key].userLastLogin);
})
}

Answer 3

看看这个：

$.ajax({

    url: 'authenticate2.php',
    type: 'POST',
    success: function(res) {
          var response = $.parseJSON(res);
          console.log(res.userLastLogin);

    })

或者你可以在你的ajax成功中做到这一点：

$.each(JSON.parse(res), function() {

  console.log(this[Object.keys(this)[6]]);

});

Answer 4

试试这个

    var json = '{"id":"1","userName":"admin","userPassword":"admin","userEmail":"admin@admin.com","userRegistrationIP":"","registrationDateTime":"2015-05-28 21:22:54","userLastLogin":"2015-05-28 21:22:54"}'; 
    for(var i = 0; i < json.length; i++) {
    var obj = json[i];

    console.log(obj.userLastLogin);
}

Answer 5

如果您只想使用userLastLogin数据，那么你可以试试这个，

while($r = mysqli_fetch_assoc($result)) {
    $rows['userLastLogin'] = $r['userLastLogin'];
}

处理返回的JSON数据

5 个答案: