如何處理 SQLSTATE[23000]：違反完整性約束：1062

Question

一切都在我的代碼中工作，但是如果您要創建一個用戶，即使您使用相同的電子郵件創建多個用戶，它也會接受我已經嘗試搜索一些解決方案並將其放入我的代碼中，但我仍然無法使其工作並且我怎樣才能使 SQLSTATE[23000]: Integrity constraint conflict: 1062 更改為“錯誤電子郵件已經存在”？ 有可能嗎？

添加表單

如果有重復，我還想顯示錯誤消息

<style type="text/css">
#dis{
display:none;
}
</style>




<div id="dis">
<!-- here message will be displayed -->
</div>


 <form method='post' id='emp-SaveForm' action="#">

<table class='table table-bordered'>

    <tr>
        <td>First Name</td>
        <td><input type='text' name='fname' class='form-control'  required /></td>
    </tr>

    <tr>
        <td>Last Name</td>
        <td><input type='text' name='lname' class='form-control'  required></td>
    </tr>

    <tr>
        <td>Contact Number</td>
        <td><input type='number' name='contactnum' class='form-control'  required></td>
    </tr>

     <tr>
        <td>Email</td>
        <td><input type='email' name='email' class='form-control' required /></td>
    </tr>

     <tr>
        <td>Password</td>
        <td><input type="password" name='pass' class='form-control'  required /></td>
    </tr>

     <tr>
        <td>Language</td>
        <td><input type='text' name='lang' class='form-control'  required /></td>
    </tr> 
    <tr>
        <td colspan="2">
        <button type="submit" class="btn btn-primary" name="btn-save" id="btn-save">
        <span class="glyphicon glyphicon-plus"></span> Save this User
        </button>  
        </td>
    </tr>

</table>

創建.php

<?php
require_once 'dbconfig.php';


if($_POST)
{
    $fname = $_POST['fname'];
    $lname = $_POST['lname'];
    $contactnum = $_POST['contactnum'];
    $email = $_POST['email'];
    $pass = $_POST['pass'];
    $lang = $_POST['lang'];
    try{

        $stmt = $db_con->prepare("INSERT INTO tbluser(fname,lname,contactnum,email,pass,lang) VALUES(:ufname,:ulname,:ucontact,:uemail,:upass,:ulang)");
        $stmt->bindParam(":ufname", $fname);
        $stmt->bindParam(":ulname", $lname);
        $stmt->bindParam(":ucontact", $contactnum);
        $stmt->bindParam(":uemail", $email);
        $stmt->bindParam(":upass", $pass);
        $stmt->bindParam(":ulang", $lang);
        if($stmt->execute())
        {
            echo "Successfully Added";
        }
        else{
            echo "Query Problem";
        }   
    }
    catch(PDOException $e){
        echo $e->getMessage();
    }
}

?>

我只是 php 新手我在數據庫中使用 PDO 不知道它是否是一件好事，感謝您的幫助，請讓我知道我是否可以使用或只是復制粘貼您的代碼:)

Answer 1

你應該編輯你的create.php並從數據庫中添加電子郵件檢查代碼，請檢查下面的代碼，以獲得更多的想法。

<?php
require_once 'dbconfig.php';


if ($_POST) {

    $fname = $_POST['fname'];
    $lname = $_POST['lname'];
    $contactnum = $_POST['contactnum'];
    $email = $_POST['email'];
    $pass = $_POST['pass'];
    $lang = $_POST['lang'];

    // it's good habit to validate email address always 
    // i checked here it should not be blank 
    if ($email != '') {

        // check in database email address is already taken
        $sel = $db_con->prepare("select email from tbluser where email = :email");
        $sel->execute(array(':email' => $email));
        $row = $sel->fetch();

        // if you found any row with current entered email then send it to else
        if ($sel->fetchColumn() == 0) {
            try{
                $stmt = $db_con->prepare("INSERT INTO tbluser(fname,lname,contactnum,email,pass,lang) VALUES(:ufname,:ulname,:ucontact,:uemail,:upass,:ulang)");
                $stmt->bindParam(":ufname", $fname);
                $stmt->bindParam(":ulname", $lname);
                $stmt->bindParam(":ucontact", $contactnum);
                $stmt->bindParam(":uemail", $email);
                $stmt->bindParam(":upass", $pass);
                $stmt->bindParam(":ulang", $lang);
                if($stmt->execute())
                {
                    echo "Successfully Added";
                }
                else{
                    echo "Query Problem";
                }   
            }
            catch(PDOException $e){
                echo $e->getMessage();
            }
        } else {
            echo $email . ' already taken, try other email address';
        }
    } else {
        echo 'Email address is blank';
    }
}

?>

Answer 2

您需要將 PDO 錯誤模式設置為。 例子：

/* @var $dbConn PDO */
$dbConn = new PDO("mysql:host=$dbHost;dbname=$dbName", $dbUsername, $dbPassword);
/* set the PDO error mode to exception */
$dbConn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

然后將您的執行包裝在 try catch 中以檢測錯誤。