Question

输入表单代码是

<form method="post" action="display.php" id="register-form" name="register-form" onsubmit="function() return false;">

<center> First Name :<pre>   <input type = "text" name="firstname" value="" ></pre><br>
Last Name:<pre><input type="text" name = "lastname" value="" ></pre><br>
State:<pre><input type="text" name = "state" value="" ></pre><br>
City:<pre><input type="text" name = "city" value="" ></pre><br>
Mobile No:
<pre>
<input type="tel" id = "mobileno" name = "mobileno" data-validation="number" value=""  ></pre><br>
Gender :<pre><ul type="none">
<li>Male:   <input type="radio" name="icheck" value="Male">
Female: <input type="radio" name="icheck" value="Female" checked></li></ul></pre>

<input type="submit" name="add" value="ADD" formaction="add.php">&nbsp;
<input type="submit" name="delete" value="DELETE">&nbsp;
<input type="submit" name= "update" value="UPDATE" formaction="display.php">&nbsp;</center>
</form>

add.php（file）（从表单中插入值）

  include("dbconnect.php");//database connect
   $firstname =$_POST['firstname'];
   $lastname = $_POST['lastname'];
   $state = $_POST['state'];
    $city = strip_tags($_POST['city']);
    $gender = strip_tags($_POST['icheck']);
    $mobileno = (int) $_POST['mobileno'];

$query = "INSERT INTO user(LastName,FirstName,State,City,MobileNo,Gender) VALUES(?,?,?,?,?,?)";
$stmt=mysqli_prepare($dbc,$query) or die(mysqli_error());
$bind='ssssis';
mysqli_stmt_bind_param($stmt,$bind,$lastname,$firstname,$state,$city,$mobileno,$gender);
mysqli_stmt_execute($stmt);

if (mysqli_stmt_affected_rows($stmt) ==
1) {
echo '<p>Your message has been
posted.</p>';
 } else {
 echo '<p style="font-weight: bold;
color: #C00">Your message could not 
be posted.</p>';
 echo '<p>' . mysqli_stmt_error($stmt) .
'</p>';
 }

 // Close the statement:
 mysqli_stmt_close($stmt);
mysqli_close($dbc);

为mobileno输入的值是7878787878，在数据库中它显示为2147483647 第二次我输入的值是8787878787，它显示相同的值。

因此代码仅显示dimatch手机号码的错误。谢谢!!

Answer 1

参数不匹配并删除手机号码前面的int

$query = "INSERT INTO user(LastName,FirstName,State,City,MobileNo,Gender) VALUES(?,?,?,?,?,?)";
    mysqli_stmt_bind_param($stmt, $bind, $lastname, $firstname, $state, $city, $mobileno, $gender);

Answer 2

更改

$bind = 'ssssis';

到

$bind = 'ssssds';

参数号无效：绑定变量的数量与php中数据库中的数字不匹配

2 个答案: