使用PHP/MySQLi查询的Ajax调用返回错误结果

问题描述

我正在尝试检查数据库中是否已存在输入值.

I am trying to check if an input value already exists in my database.

到目前为止，我得到了以下内容，但这始终使我返回无效"(即使我输入了匹配的值)，所以我的猜测是我的SELECT查询后的部分或If Else逻辑中存在错误.

So far I got the following but this always returns me "Invalid" (even if I enter a matching value) so my guess is I have an error in the part after my SELECT query or in my If Else logic.

有人可以告诉我该怎么做吗?

Can someone tell me how to do this right?

我的JS:

$('#btnCheck').click(function() {
    var username = $('#username').val();

    $.ajax({
        url: 'userSubmit.php',
        type: 'POST',
        data: {username: username},
        success: function(response){
            if(response == 'valid') {
                alert('Valid');
            } else {
                alert('Invalid');
            }
        }
    });
});

我的PHP:

$username = $_POST['username'];
$conn = new mysqli($host, $username, $password, $database);
if($conn->connect_error) {
    die("Connection Error: " . $conn->connect_error);
}
$stmt = $conn->prepare("SELECT username FROM users WHERE username = ?");
$stmt->bind_param("s", $username);
$stmt->execute();

if($stmt->num_rows > 0) {
    echo "valid";
} else {
    echo "invalid";
}

推荐答案

在调用num_rows之前，您需要调用store_result()以便实际记录行数:

You need to call store_result() in order for the number of rows to actually be recorded before calling num_rows:

...
$stmt->execute();
$stmt->store_result();

if($stmt->num_rows > 0) {
    ...

必须调用store_result()的原因是，如果没有此调用，则实际上不会获取任何结果，因此行数(num_rows)将是0.

The reason why you must call store_result() is that without this, no results have actually been fetched, and the number of rows (num_rows) will therefore be 0.

参考num_rows的官方文档:

与此相关的链接是此处.

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