Question

您好，我对编程有点陌生，我正尝试使用ajax，并希望从此php中获取值，但无法使其正常工作

$userID = $_SESSION['id'];

$query = "SELECT * from ipcr where userID = '".$userID."'";
$result = mysqli_query($conn, $query);

while ($row = mysqli_fetch_assoc($result)) {
    $current_id = $row['id'];
    $current_details = $row['details'];
    $current_dateCreated = $row['dateCreated'];
    $current_ipcrCode = $row['ipcrCode'];
    $current_employeeNumber = $row['employeeNumber'];

    $array = array(
        'id'=>$current_id,
        'details' => $current_details,
        'dateCreated' => $current_dateCreated,
        'ipcrCode' => $current_ipcrCode,
        'employeeNumber' => $current_employeeNumber
    );

    echo json_encode($array);
}

但我不断收到错误消息：

SyntaxError：JSON中位置173上的意外令牌<< / p>

当我尝试验证我的json时 this error

这是php回显的实际输出。

{"id":"21836","details":"Details here","dateCreated":"2018-08-01 14:25:28","ipcrCode":"22703","employeeNumber":"140010663"}
{"id":"21837","details":"details here","dateCreated":"2018-08-01 14:25:57","ipcrCode":"22703","employeeNumber":"140010663"}

我使用json_encode的方式有问题吗？看来我呼应的格式是错误的。

这也是我的脚本的样子

function get_ipcr() {
    var userID = <?php echo $_SESSION['id']; ?>;

    $.ajax({
        type: "POST",
        url: "../includes/php/load_ipcr.php",

        dataType: "json",
        success: function(results) {
            alert("success");
        },
        error: function (jqXHR, textStatus, errorThrown) {
            alert(errorThrown);
        }

    });

}

Answer 1

问题是您正在循环回显JSON对象。这样会导致{valid JSON}{valid JSON}之类的东西，它不是有效的JSON。

您可以通过以下方式解决此问题：

$employees = array();
while ($row = mysqli_fetch_assoc($result)) {
    $current_id = $row['id'];
    $current_details = $row['details'];
    $current_dateCreated = $row['dateCreated'];
    $current_ipcrCode = $row['ipcrCode'];
    $current_employeeNumber = $row['employeeNumber'];
// Record into employees array.
    $employees[] = array(
        'id'=>$current_id,
        'details' => $current_details,
        'dateCreated' => $current_dateCreated,
        'ipcrCode' => $current_ipcrCode,
        'employeeNumber' => $current_employeeNumber
    );
}
echo json_encode($employees);

Answer 2

欢迎来到stackoverflow！

您将要从不同的评论/答案中听到几件事，我建议您听听他们的声音-似乎不堪重负，或者像“信息太多”一样-但请相信我，我们都在这里为您提供帮助！

问题的答案
将 entire JSON字符串烘烤到单个数组（数组的数组）中，而不是在循环中多次回显JSON字符串。

其他提示
另外，作为提示，通过分配所有变量（$current_id = $row['id']等），您可以自己进行更多输入。您可以直接直接使用$row['id']……在下面，您可以看到我展示了一种简化代码的方法：

// shorthand version of $array = array()
$array = [];
while ($row = mysqli_fetch_assoc($result)) {
    // don't bother assigning $row values to a series of variables
    // removed all the lines like the next one....
    // $current_id = $row['id'];
    // push this record onto the array of arrays....
    $array[] = [
        // assign the array values directly from $row
        'id'             =>$row['id'],
        'details'        => $row['details'],
        'dateCreated'    => $row['dateCreated'],
        'ipcrCode'       => $row['ipcrCode'],
        'employeeNumber' => $row['employeeNumber']
    ];
}

echo json_encode($array);

最后，最重要
请，请-不要写这样的查询语句。它对SQL注入攻击开放，这是一个严重的问题。

请see this post了解如何prevent SQL injection attacks。

无法使用Ajax从JSON获取多个值

2 个答案: