警告:mysqli_error()期望恰好有1个参数，在register.php文件中给出的0个参数 [英] Warning: mysqli_error() expects exactly 1 parameter, 0 given in my register.php file

问题描述

我严格按照w3schools.com中的代码进行操作..但仍然出现上述错误，并且数据库表也未更新.

I followed the code exactly as in w3schools.com..but still I am getting the above error and the database table is also not updated.

我的html和php代码如下:

My html and php code is as follows :

<form action="register.php" method="post">
<select name="title" size="1">
  <option>Mr.</option>
  <option>Ms.</option>
  <option>Prof.</option>
  <option>Dr.</option>
</select>
<label class="label">Given Name: <input name="givenname" type="text" value="Enter Your First Name" maxlength="30" /></label>
<label class="label">Surname: <input name="surname" type="text" /></label>
<label class="label">Address: <input name="address" type="text" maxlength="80" /></label>
<label class="label">Phone No: <input name="phoneno" type="text" /></label>
<label class="label">Email ID: <input name="emailid" type="text" /></label>
<label class="label">Fax: <input name="fax" type="text" /></label>
<label class="label">Pincode: <input name="pincode" type="text" /></label>
<label class="label">Country: <input name="country" type="text" /></label>
<input name="submit" type="submit" />
</form>

<?php
$con=mysqli_connect("my_Ipaddress","abcd","abcd");
//database connection
    if (mysqli_connect_errno()) 
    {
    echo "Failed to connect to Database...There must be some error! We are working on it!!" . mysqli_connect_error();
    }

    $sql="INSERT INTO Registrations (Title, Given Name, Surname, Address, Phone No, Email ID, Fax, Pincode, Country)
    VALUES ('$_POST[title]','$_POST[givenname]','$_POST[surname]','$_POST[address]','$_POST[phoneno]','$_POST[emailid]','$_POST[fax]','$_POST[pincode]','$_POST[country]')";
    if (!mysqli_query($con,$sql))
    {
    die('Error: ' . mysqli_error());
    }
    echo "1 record added";
    mysqli_close($con);
    ?>

推荐答案

我收到了您的错误，这里是

I got your error, here it is

mysqli_error()语法错误.

mysqli_error() syntax is wrong.

更改

die('Error: ' . mysqli_error());

到

die('Error: ' . mysqli_error($con));

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