我是
PHP OOP的新手,所以这个问题一定很愚蠢.
我无法通过PHP创建SQL查询.我已多次阅读代码,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.
我无法通过PHP创建SQL查询.我已多次阅读代码,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>
现在来到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; ?>
Object of class MysqLi_stmt could not be converted to string
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半部分,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.
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 ;
.....
