Question

我是PHP OOP的新手，所以这个问题必须非常愚蠢。我无法通过PHP创建SQL查询。我已多次阅读该代码，但我无法找到任何差异，甚至编辑器也没有显示任何错误。我使用的是PHP 5.5.13，MYSQL 5.5.24和APCHE Server 2.2.22。

以下是代码：

Test_signup.php

<!DOCTYPE HTML>
    <html>
        <head>
            <title>
            Test Sign Up
            </title>
        </head>
        <body>
                <form action = "Signup.php" method = "POST" name = "test_signup">
                Full Name: <input type = "text" name = 'full_name'>
                User Name: <input type = 'text' name = 'user_name'>
                Email: <input type = 'text' name = 'email_add'>
                <input type = "submit" name = "submit">
            </form>

        </body>
    </html>

现在进入 Signup.php

<?php
$con = new mysqli('localhost', 'root', '', 'my_database');
      if ($con->connect_error)
      {
        echo 'Failed to connect' . $con->connect_error;
      }
      else
      {
        echo 'Connected';
        $stmt_chk_email = $con->prepare('SELECT * FROM `user_information` WHERE `Email` = ?');
        $stmt_chk_email->bind_param('s', $_POST['email_add']);
        echo $stmt_chk_email;
?>

尝试运行此代码时，我收到错误：

类mysqli_stmt的对象无法转换为字符串

当然原始查询比这里发布的要大得多。我已经编辑了查询的后面部分，因为我发现问题存在于SQL“SELECT”语句中，但我无法弄明白。请帮助我。谢谢。

Answer 1

你不能回应这个

~~echo $ stmt_chk_email;~~

你可能想要回应这个

     $stmt_chk_email = $con->prepare('SELECT column1 ,column2,... FROM `user_information` WHERE `Email` = ?');
     $stmt_chk_email->execute();
     $stmt_chk_email->store_result();
     $stmt_chk_email->bind_result($column1 ,$column2,.....);
     $stmt_chk_email->fetch();
  echo $column1;
  echo $column2 ;
  .....

Answer 2

        $stmt_chk_email = $con->prepare('SELECT `column` FROM `user_information` WHERE `Email` = ?');
        $stmt_chk_email->bind_param('s', $_POST['email_add']);
        $stmt_chk_email->execute();
        $stmt_chk_emeil->store_result();
        $stmt_chk_email->bind_result($check_email);
        $stmt_chk_email->fetch();
        echo $check_email;

OR

    $stmt_chk_email = $con->prepare('SELECT * FROM `user_information` WHERE `Email` = ?');
    $stmt_chk_email->bind_param('s', $_POST['email_add']);
    $stmt_chk_email->execute();

    $result = $stmt_chk_email->get_result();

    if($result->num_rows == 1)
    {
         $result = $result->fetch_array();
         echo $result['column_name'];
    }

可捕获的致命错误：类mysqli_stmt的对象无法转换为字符串

2 个答案: