Question

尝试将变量 $ user_id 绑定到准备好的语句，该数组作为jsonarray发送到Java文件。以下代码可以正常工作，但是它没有绑定参数，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'],
            )
            );
        }

下面的代码包含与绑定参数有关的错误：

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：
删除了占位符的引用。
$ mysql_qry =“从user_qualification中选择*，其中qual_id =？”；

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

编辑3：
在if语句后 var_dump 变量$ user_id
object（mysqli_result）＃3（5）{[“” current_field“] => int（0）[” field_count“] => int（1）[” lengths“] => NULL [” num_rows“] => int（1 ）[“ type”] => int（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();
?>

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

编辑6：
现在可以正常工作了：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

共享instance.get({ plain: true })的转储后，似乎是mysql结果。为了从那里提取ID，您应该使用$user_id。

mysqli_fetch_assoc

我假设您将查询用作：“ SELECT ID From ...”，就像您获得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);

PHP-将变量绑定到Prepared语句时出错

1 个答案: