Question

这是我在尝试以JSON格式编码字符串时获得结果的方式。我希望该网址应该在：http://ipaddress/Up/user_images/549279.jpg

结果：

{＆＃34;导致＆＃34 ;: [{＆＃34; ID＆＃34;：＆＃34; 42＆＃34;＆＃34;名称＆＃34;：＆＃34; HTTP：// IPADDRESS /上/ user_images / 380289.jpg＆＃34;}，{＆＃34; ID＆＃34;：＆＃34; 43＆＃34;＆＃34;名称＆＃34;：＆＃34; HTTP：// IPADDRESS /上/ user_images / 995062.jpg＆＃34;}，{＆＃34; ID＆＃34;：＆＃34; 44＆＃34;＆＃34;名称＆＃34;：＆＃34; HTTP：// IPADDRESS /上/ user_images / 171544.jpg＆＃34;}，{＆＃34; ID＆＃34;：＆＃34; 41＆＃34;＆＃34;名称＆＃34;：＆＃34; HTTP：// IPADDRESS / Up / user_images / 549279.jpg＆＃34;}]}

警告：json_decode（）期望参数1为字符串，第26行的C：\ xampp \ htdocs \ jsn \ PHP_Scripts \ getAllEmp.php中给出的数组

//creating a blank array 
$result = array();
$r1 = array();

//looping through all the records fetched
while ($row = mysqli_fetch_array($r)){
    //Pushing name and id in the blank array created 
    array_push($result,array(
        "id"=>$row['userID'],
        "name"=>$ur.$row['userPic'],
    ));
} 

//Displaying the array in json format 
echo json_encode(array('result'=>$result));
$r1= json_decode($result, true);
echo ($r1);
//  echo json_decode(array("result"=>$r1));

mysqli_close($con);

Answer 1

尝试以下代码

    //creating a blank array 
    $result = array();
    $r1=array();
    //looping through all the records fetched
    while($row = mysqli_fetch_array($r)){
        //Pushing name and id in the blank array created 
            array_push($result,array(
            "id"=>$row['userID'],
            "name"=>$ur.$row['userPic'],
        ));
    } 
    //Displaying the array in json format 
    $jsonData =  json_encode($result);
    echo $jsonData;  // **MODIFIED LINE**
    $r1 = json_decode($jsonData, true); // **MODIFIED LINE**
    print_r($r1);
    mysqli_close($con);

带有图像URL的Json字符串或用于url的Json格式编码

1 个答案: