Question

我需要返回一个 geojson 数据来响应ajax请求。

这是我的ajax代码：

postData="{\"Title\":\""+title.getText().toString()
            +(getIntent().hasExtra(Common.REQID)?"\",\"ID\":\""+getIntent().getExtras().getString(Common.REQID):"")//checking update or insert
            +"\",\"ReqDateFormatted\":\""+date.getText().toString()
            +"\",\"ReqForCompany\":\""+companyList.get(companySpinner.getSelectedItemPosition())[0]//Company code
            +"\",\"RequisitionBy\":\""+employeeList.get(employeeSpinner.getSelectedItemPosition()==0? -1:employeeSpinner.getSelectedItemPosition()-1)
            +"\",\"ReqStatus\":\"Open\"" +
            ",\"RequisitionDetailList\":" +
             detailItemsJSON
            +",\"userObj\":{\"UserName\":\""+userName+"\"}}";

这是 database_query.php 由ajax触发的php文件：

$.ajax({
    url : 'database_query.php',               
    data:'begin='+$("#begin").val(),
    type : 'GET',
    dataType : 'json',
    success : function(code_json, statut){
        alert("success" +" "+code_json+" "+statut)
    },
    error : function(resultat, statut, erreur){
        alert(erreur)
    }
});

我得到的错误信息：

<?php
try{
    // On se connecte à MySQL

    $bdd = new PDO('mysql:host=localhost;dbname=nyc;charset=utf8', 'root', '');
} catch(Exception $e) {
    // En cas d'erreur, on affiche un message et on arrête tout
    die('Erreur : '.$e->getMessage());
}
if(isset($_GET["begin"])){
    $data={"features": [{"geometry": {"coordinates": [-73.95240784, 40.81072617], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.96302032, 40.71183395], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.84346008, 40.75595093], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.92481232, 40.75424576], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.92950439, 40.75645065], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.9250412, 40.76170349], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.90320587, 40.74570465], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.84425354, 40.72134018], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.98892975, 40.70240021], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.95552826, 40.8044548], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.95846558, 40.71715546], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.9287262, 40.75831223], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.90457916, 40.74129486], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.94830322, 40.80869675], "type": "Point"}, "properties": {}, "type": "Feature"}, {"geometry": {"coordinates": [-73.93263245, 40.8042717], "type": "Point"}, "properties": {}, "type": "Feature"}], "type": "FeatureCollection"}
    echo $data;
}

?>

我想念或弄错了什么？

Answer 1

要返回json作为响应，请不要手动创建它，而是创建一个普通的php数组并执行

echo json_encode($data);

将json返回到您的ajax调用

Answer 2

json_encode也有few optoins for encoding，这使您免于头痛。手工字符串中的问题很可能是您使用了句点（在编码奇怪的事情发生时）。您可以使用类似的方法来确保浮点数正确解析：

echo json_encode($result, JSON_PRETTY_PRINT | JSON_UNESCAPED_UNICODE | JSON_NUMERIC_CHECK | JSON_UNESCAPED_SLASHES);

如何从PHP返回JSON数据

2 个答案: