PHP - 将变量绑定(bind)到准备好的语句时出错

Question

尝试将变量 $user_id 绑定(bind)到准备好的语句，数组作为 jsonarray 发送到 java 文件。下面的代码工作得很好，但它没有绑定(bind)参数，其中 qual_id 的值是静态的。

if ($user_id->num_rows >= 1) {
        $mysql_qry = "select * from user_qualification where qual_id='1'";
        //$stmt = mysqli_stmt_init($conn);
        //$result = mysqli_stmt_prepare($stmt, $mysql_qry);
        //mysqli_stmt_bind_param($result, "i", $user_id);

       $result = mysqli_query($conn,$mysql_qry);
        //mysqli_stmt_execute($stmt);

        $data_item = array();
        while($row = mysqli_fetch_assoc($result)){
            array_push($data_item, 
            array('u_school'=>$row['school'],
            'u_hschool'=>$row['hschool'],
            'u_undergrad'=>$row['ugrad'],
            'u_grad'=>$row['grad'],
            'u_phd'=>$row['phd'],
            )
            );
        }

以下代码包含与绑定(bind)参数有关的错误:

if ($user_id->num_rows >= 1) {
    $mysql_qry = "select * from user_qualification where qual_id='?'";
    $stmt = mysqli_stmt_init($conn);
    $result = mysqli_stmt_prepare($stmt, $mysql_qry);
    mysqli_stmt_bind_param($result, "s", $user_id);

    //$result = mysqli_query($conn,$mysql_qry);
    mysqli_stmt_execute($stmt);

    $data_item = array();
    while($row = mysqli_fetch_assoc($result)){
        array_push($data_item, 
        array('u_school'=>$row['school'],
        'u_hschool'=>$row['hschool'],
        'u_undergrad'=>$row['ugrad'],
        'u_grad'=>$row['grad'],
        'u_phd'=>$row['phd'],
        )
        );
    }

qual_id 列也是 bigint。

编辑 1:
去掉占位符的引号。
$mysql_qry = "select * from user_qualification where qual_id=?";

编辑 2 :
从“s”更改为“i”。
mysqli_stmt_bind_param($result, "i", $user_id);

编辑 3 :
var_dump 变量 $user_id 在 if 语句之后
object(mysqli_result)#3 (5) { ["current_field"]=> int(0) ["field_count"]=> int(1) ["lengths"]=> NULL ["num_rows"]=> int(1 ) ["类型"]=> 整数(0) } []

编辑 4 :
php文件

<?php
require "conn.php";
$user_name ="omx123"; //$_POST["user_name"];
if ($mysql_qry = $conn->prepare("Select id from UserLoginDetails where username=?")) {
    $mysql_qry->bind_param("s", $user_name);
    $mysql_qry->execute();
    $user_id = $mysql_qry->get_result();

    if ($user_id->num_rows >= 1) {
        var_dump($user_id);
        $mysql_qry = "select * from user_qualification where qual_id=?";
        $stmt = mysqli_stmt_init($conn);
        $result = mysqli_stmt_prepare($stmt, $mysql_qry);
        mysqli_stmt_bind_param($result, "i", $user_id);

        //$result = mysqli_query($conn,$mysql_qry);
        mysqli_stmt_execute($stmt);

        $data_item = array();
        while($row = mysqli_fetch_assoc($result)){
            array_push($data_item, 
            array('u_school'=>$row['school'],
            'u_hschool'=>$row['hschool'],
            'u_undergrad'=>$row['ugrad'],
            'u_grad'=>$row['grad'],
            'u_phd'=>$row['phd'],
            )
            );
        }
        echo json_encode($data_item);
    }
} 
$conn->close();
?>

编辑 5:
感谢您的帮助用户:dWinder
$user_id 上的 var_dump :
整数(1)[]
但是数组仍然是空的。

编辑 6:
现在工作得很好 :D

$mysql_qry = "select * from user_qualification where qual_id=?";
         $stmt = mysqli_stmt_init($conn);
         mysqli_stmt_prepare($stmt, $mysql_qry);
         mysqli_stmt_bind_param($stmt, "i", $user_id_int);
         mysqli_stmt_execute($stmt);
         $result=mysqli_stmt_get_result($stmt);

Answer 1

在您共享 $user_id 的转储后，它似乎是 mysql 结果。为了从那里提取您的 ID，您应该使用 mysqli_fetch_assoc。

if ($user_id->num_rows >= 1) {
    $row = mysqli_fetch_assoc($user_id)
    $user_id_int = $row["id"]; // or what ever you used to call it

    // now you can call the bind...
    $mysql_qry = "select * from user_qualification where qual_id=?";
    $stmt = mysqli_stmt_init($conn);
    $result = mysqli_stmt_prepare($stmt, $mysql_qry);
    mysqli_stmt_bind_param($result, "i", $user_id_int);

我假设您在获取 $user_id

时将查询用作:“SELECT id From ...”