当stmt - ＆gt; execute（）执行时，无法跟踪错误的确切位置

Question

我有一个名为personal_detail的表。表单用于使用用户创建的条目填充此表。查询未成功执行，我无法在查询中找到错误的确切位置。按下提交按钮后我收到的消息未创建。这是由我在php文件的末尾提供的。如果stmt - ＆gt; execute（）返回false。

这是personal_detail表结构：

米/ BvbLS.jpg

这是表格代码：

<form action="personal.php" method="post">
    Name: <input name="name" type="text" size="20" maxlength="40" ><br>
    CNIC : <input name="cnic" type="text" size="20" maxlength="15"><br>
    Date: <input name="booking-date" type="date" size="20" ><br>
    Ocassion: <select name="ocassion" size="1">
        <option value="">Barat</option>
        <option value="">Walima</option>
        <option value="">option3</option>
        <option value="">option4</option>
        <option value="">option5</option>
        <option value="">option6</option>
    </select><br>
    Address:<input name ="address" type="text" size="20" maxlength="50"><br>
    Phone-No:<input name="phone-no" type="text" size="20" maxlength="11"><br>
    Bride-Mobile:<input name="bride-mobile" size="20" maxlength="11"><br>
    Groom-Mobile:<input name="groom-mobile" size="20" maxlength="11"><br>
    Family-Mobile:<input name="family-mobile" size="20" maxlength="11"><br>
    E-mail:<input name="email" type="text" size="20" maxlength="30"><br>
    Who may i thank for refering you?:<input name="refering" type="text" size="20" maxlength="40"><br>
    Do you provide consent to share images on our official web page:<br>
    <input type="radio" name="share" value="Yes">Yes <br>
    <input type="radio" name="share" value="No">No<br>
    If yes:<br>
    With identity: <br><input type="radio" name="permission" value="Yes">Yes<br>
    <input type="radio" name="permission" value="No">No<br>
    <input class="btn btn-primary btn-large" type="submit" value="Submit" name="Submit-Personal">
    <input class="btn btn-primary btn-large" type="reset" value="Reset">

这是php文件：

$server="localhost";
$user="root";
$password="";
$database="camouflage_studio";

$con = mysqli_connect($server,$user,$password,$database);
if (mysqli_connect_errno()){
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// prepare and bind
$stmt = $con->prepare("insert into personal_detail (Name, CNIC, Date, Ocassion, Address, Phone_No, Bride_Mobile, Groom_Mobile, Family_Mobile,EMail,Referring,Share,Permission) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)");

$stmt->bind_param("ssssssiiissss", $name, $cnic, $date, $ocassion, $daddress, $phoneno, $bridemobile, $groommobile, $familymobile, $email, $refering, $share, $permission);

$name = $_POST['name'];
$cnic = $_POST['cnic'];
$date = $_POST['booking-date'];
$ocassion = $_POST['ocassion'];
$address = $_POST['address'];
$phoneno = $_POST['phone-no'];
$bridemobile = $_POST['bride-mobile'];
$groommobile = $_POST['groom-mobile'];
$familymobile = $_POST['family-mobile'];
$email = $_POST['email'];
$refering = $_POST['refering'];
$share = $_POST['share'];
$permission = $_POST['permission'];

$stmt->execute();
$result = mysqli_stmt_execute($stmt);
if($result)
    echo "New records created successfully";
else
    echo "not created";

// close connection
$stmt->close();
$con->close();

Answer 1

这是一个代码块，可以执行完整的错误检查。

• 您使用了程序和面向对象语法的混合;我把所有东西都设置为面向对象。
• 我已经对表名和所有列进行了反向排列（尽管我只是建议Date进行反对。
• 我已将所有mysql关键字大写。
• 我已经删除了只传递$_POST到bind_param()的值的不必要的变量声明。
• execute()一直隐藏在以bind_param()开头的条件结束时。原因/结果是，如果bind_param()返回false，则甚至不会调用execute()，并且会显示错误消息。
• 当然，当您的代码“上线”时，请确保不要回复任何这些错误消息 - 只在调试时显示它们。

if(!$con=new mysqli($server,$user,$password,$database)){
    echo "Connection Error: ",$con->connect_error;  // do not echo when live
}elseif($stmt=$con->prepare("INSERT INTO `personal_detail` (`Name`,`CNIC`,`Date`,`Ocassion`,`Address`,`Phone_No`,`Bride_Mobile`,`Groom_Mobile`,`Family_Mobile`,`EMail`,`Referring`,`Share`,`Permission`) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)")){
    if($stmt->bind_param('ssssssiiissss',$_POST['name'],$_POST['cnic'],$_POST['booking-date'],$_POST['ocassion'],$_POST['address'],$_POST['phone-no'],$_POST['bride-mobile'],$_POST['groom-mobile'],$_POST['family-mobile'],$_POST['email'],$_POST['refering'],$_POST['share'],$_POST['permission']) && $stmt->execute()){
        echo "New record created successfully"; // New ID# is ",$stmt->insert_id;
    }else{
        echo "New record not created";
        echo "Statement Error: ",$stmt->error;  // do not echo when public
    }
    $stmt->close(); 
}else{
    echo "Prepare Error: ",$con->error;  // do not echo when public
}
$con->close();