在jQuery AJAX Success中獲取MySql的Specfic響應

Question

好吧，我有這個ajax代碼，它將從Success塊中的MySql返回結果。

$.ajax({
   type:"POST",
   url:"index.php",
   success: function(data){
            alert(data);                
        }
});

我的查詢

$sql = "SELECT * FROM tablename";
$rs=parent::_executeQuery($sql);
$rs=parent::getAll($rs);
print_r($rs);
return $rs

我的響應數組在成功AJAX的警報中

Array
(
    [0] => Array
        (
            [section_id] => 5
            [version] => 1
            [section_name] => Crop Details
            [id] => 5
            [document_name] => Site Survey
            [document_master_id] => 1
            [document_section_id] => 5
        )

    [1] => Array
        (           
            [section_id] => 28
            [version] => 1
            [section_name] => Vegetative Report           
            [id] => 6
            [document_name] => Site Survey
            [document_master_id] => 1
            [document_section_id] => 28
        )

)

我想從結果中只獲取section_name和document_name ，以便我可以將這兩個值附加到我的列表中。

Answer 1

不要使用print_r()返回響應，請使用json_encode() ：

echo json_encode($rs);

然后在Javascript中，你可以這樣做：

$.ajax({
   type:"POST",
   url:"index.php",
   dataType: 'json'
   success: function(data){
        for (var i = 0; i < data.length; i++) {
            console.log(data[i].section_name, data[i].document_name);          
        }
    }
});

Answer 2

使用更改您的選擇查詢

$sql = "SELECT section_name,document_name FROM tablename";

Answer 3

$sql = "SELECT * FROM tablename";
$rs=parent::_executeQuery($sql);
$rs=parent::getAll($rs);

return json_encode($rs);

的index.php

$.ajax({
   type:"POST",
   url:"index.php",
   success: function(data){
        alert(data);
        var data = JSON.parse(data);
        /* you can use $.each() function here  */                
    }
});

Answer 4

做這個：

$sql = "SELECT * FROM tablename";
$rs=parent::_executeQuery($sql);
$rs=parent::getAll($rs);
$resp = array();
foreach( $rs as $each ){
    $resp[]['section_name'] = $each['section_name'];
    $resp[]['document_name'] = $each['document_name'];
}
return json_encode($resp);

像這樣訪問JSON響應：

for (var i = 0; i < data.length; i++) {
    console.log(data[i].section_name, data[i].document_name);          
}