Question

我有一个html表单中的两个组合框，它们将使用jQuery和PHP从我的数据库中动态填充。当其中一个组合框被更改时，我调用我的jQuery，并将更改的组合框的值发送到我的php页面，以便从我的数据库中查找值。

function valueChanged(department){

        $.get("AdminAjax.php",$("#department").serializeArray(),

        function(data){
            alert(data);//debug code that ouputs the returned value
            var i =0;
            for(i=0; i < data.length; i++){
                //create a string that will be appended into the second combo box.   
                var option = '<option value="' + data.Category[i] +               
                '">=' + data.Category[i] + '</option>';
                $("#category").append(option);
            }
        },"html" //debug for the return
        );
    }

我知道组合框的值是基于php页面的试错而传递的。

<?php
$department = mysql_real_escape_string($_GET["department"]);
//$department is the value passed from the jQuery.

$categorySQL = "Select Category from pagedetails where Department = '$department'";
//get the values for the second combo box based on the department

$rs = mysql_query($categorySQL);//<---this is where it fails.
//I have echoed out $rs and $rowCategory after I have fetched it. 
//They return Resource #4 and Array respectively.

while($rowCategory = mysql_fetch_assoc($rs)){
//I am expecting multiple records to be returned.
   $json_out = json_encode($rowCategory);
}
echo $json_out;
?>

Answer 1

你的回答错误，你的php是错误的，你需要每次while为真时使用$.getJSON或$.ajax而不是$.get. You are resetting the $ json_out`变量。您应该将所有值保存到数组中，然后对其进行json_encode一次。这也将确保你的json成功是有效的json。

试试这个：

$json_out = array();
while($rowCategory = mysql_fetch_assoc($rs)){
//I am expecting multiple records to be returned.
   $json_out[] = $rowCategory;
}
echo json_encode($json_out);

Answer 2

您在每次抓取时都会覆盖$json_out。

而是尝试：

$json_out = array();
while($rowCategory = mysql_fetch_assoc($rs)){
   $json_out[] = $rowCategory;
}
echo json_encode($json_out);

AJAX：没有从数据库获取关联数组

2 个答案: