Question

我在PHP中有下一个代码：

    $servername = "***";
    $username = "***";
    $password = "***";
    $dbname = "***";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    $sql = "select * from `opencartdb`.`oc_design` where name='mydesign '";
    $res=$conn->query($sql);
    $records=$res->fetch_assoc();//success 

    $result2=$records['info'];// already json 
    $conn->close();


    echo ($result2);

我试着用下一个JS抓住它：

jQuery.ajax({
            dataType: "json",
            url: "ajax.php?type=loadDesign&user_id="+user_id+"&design_id="+key
        }).success(function( data ) {
            alert(0);
            console.log((data));
        }).always(function(){
            alert(1);
        });

最后我看到警报1 Debuger向我展示了下一个结果。回复：

{"vectors":"{"front":{"0":{"type":"text","width":"55px","height":"27px","top":"151px","left":"86px","zIndex":"1","svg":"<svg width=\"54.9375\" height=\"27.09375\" viewBox=\"0 0 54.9375 27.09375\" xmlns=\"http://www.w3.org/2000/svg\" xmlns:xlink=\"http://www.w3.org/1999/xlink\"><g id=\"0.5646970888306251\"><text fill=\"#FF0000\" stroke=\"none\" stroke-width=\"0\" stroke-linecap=\"round\" stroke-linejoin=\"round\" x=\"\" y=\"\" text-anchor=\"start\" font-size=\"24px\" font-family=\"arial\" data-textcurve=\"1\" data-itemzoom=\"1 1\" data-textspacing=\"0\"><textPath xmlns:xlink=\"http://www.w3.org/1999/xlink\" xlink:href=\"http://138.68.62.219/Design-Apple-T-shirt-PC55#textPath-item-0\"><tspan dy=\"0\">Hello</tspan></textPath></text></g><defs><path id=\"textPath-item-0\" d=\"M 0.125 22.117808976867764 A 3093.9720937064453 3093.9720937064453 0 0 1 54.124314613414626 22.117808976867764\"></path></defs></svg>","rotate":0,"text":"Hello","color":"#FF0000","fontFamily":"arial","align":"center","outlineC":"none","outlineW":0}},"back":{}}","teams":"{}","fonts":"","design_id":"1475007569235335440","image":"uploaded/2016/09/design-1475007569235335440.png","parent_id":"0","product_id":"178","product_options":"000000","title":"","description":""}

此行在调试中绘制：

<svg width=\"54.9375\...

Answer 1

将php中的响应类型设置为json。否则jQuery可能不接受响应类型并且不会触发成功函数。

header('Content-Type: application/json'); 
echo json_encode($result2);

您还应该添加一个字符集以避免BOM：

header('Content-type:application/json;charset=utf-8');

jQuery成功不会触发代码200

1 个答案: