JSON.parse在尝试解析多个JSON对象时抛出"意外令牌"错误。我以前使用过这种方法,但仅当数组包含一行查询结果时使用。
如有任何帮助,我们将不胜感激。谢谢
错误
Uncaught SyntaxError: Unexpected token {
JavaScript
if(window.XMLHttpRequest){
var request = new XMLHttpRequest();
} else {
var request = new ActiveXObject(microsoft.XMLHTTP);
}
request.open('POST', 'controllers/engineersOutput.php', true);
request.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');
request.onreadystatechange = function(){
if((request.readyState==4) && (request.status==200 || request.status==304)) {
var str = request.responseText;
var data = JSON.parse(str);
console.log(data);
}
}
request.send();
PHP
public function getEngineerDetails(){
include "connect.php";
$query = "SELECT * FROM `engineers`";
$result = mysqli_query($con, $query);
while($row=mysqli_fetch_assoc($result)){
$name = $row['name'];
$pic = $row['pic'];
$insertedEngs = array( "name"=>$name, "pic"=>$pic );
echo json_encode($insertedEngs);
}
}
看起来您的代码正在返回JSON编码对象的序列,而这对于JSON解析器来说是无法识别的。将对象收集到一个数组中(在PHP代码中),然后在数组完成时通过json_encode
回显整个数组。然后,您的客户端将收到一个JSON数组。
从服务器创建的JSON字符串很可能因为创建方式而格式错误。
不要在循环中构造JSON字符串,首先构造并完成数组,然后最后编码:
public function getEngineerDetails() {
include "connect.php";
$query = "SELECT * FROM `engineers`";
$result = mysqli_query($con, $query);
// define a container first
$data = array();
while($row = mysqli_fetch_assoc($result)){
// while inside the loop
// construct the array first
$insertedEngs = array("name" => $row['name'], "pic" => $row['pic']);
$data[] = $insertedEngs; // continually push
}
// then encode output when done
echo json_encode($data);
}