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我創建了一個小型註冊粘性表單。一切工作正常，但如果我輸入任何錯誤的值，如姓名中的數字，年齡的字母，甚至錯誤的電子郵件格式，然後數據仍保存在數據庫中，我無法弄清楚驗證問題。 另一個是搜索選項。每當我在搜索框中輸入任何名字或姓氏時，它應該顯示數據庫中的名稱或它將顯示一條錯誤消息。 任何1可以建議我該怎麼辦..請參閱下面我的代碼。PHP，MySQL驗證故障和搜索不起作用？
sticky_form代碼...
<html>
<head>
<?php
global $fname,$lname,$gender,$age,$course,$email;
if(isset($_POST['register']))
{
$fname=$_POST['fname'];
$lname=$_POST['lname'];
$gender=$_POST['gender'];
$age=$_POST['age'];
$course=$_POST['course'];
$email=$_POST['email'];
if (preg_match("/[a-zA-Z ]+$/", $_POST['fname'])) {
$fname = trim($_POST['fname']);
}
else
{
echo '<p>The First name is empty or has illegal characters! To edit please go the link Display Data Information</p>';
//$error = true;
}
if (preg_match("/[a-zA-Z ]+$/", $_POST['lname'])) {
$lname = trim($_POST['lname']);
}
else
{
echo '<p>The last name is empty or has illegal characters! To edit please go the link Display Data Information</p>';
$error = true;
}
if(isset($_POST['gender']))
{
$gender = $_POST['gender'];
}
else
{
echo "<p>No gender found!</p>";
}
if (preg_match("/[a-zA-Z ]+$/", $_POST['age'])) {
$age = trim($_POST['age']);
}
else
{
echo '<p>Please enter age. Or your age contains illegal characters</p>';
//$error = true;
}
if(isset($_POST['course']))
{
$course = $_POST['course'];
}
else
{
echo "<p>Please Select Course!</p>";
}
// Validate the email:
if (preg_match("/^[\w.-][email protected][\w.-]+\.[A-Za-z]{2,6}$/", $_POST['email'])){
$email = trim($_POST['email']);
}
else
{
echo '<p>The email is empty or has illegal characters! To edit please go the link Display Data Information</p>';
//$error = false;
}
echo "<br/>";
echo "<br/>";
echo "<br/>";
}
if($fname&&$lname&&$gender&&$age&&$email&&$course)
{
require_once('connection.php');
$query = mysql_query("INSERT INTO members SET FirstName='$fname', LastName='$lname', Gender='$gender', Age='$age', Email='$email', Course='$course'") or die(mysql_error());
if($query){
echo"Your Data Successfully Saved";
}
else
{
echo "Please recheck your Data!";
}
}
?>
</head>
<body id="body">
<h2><strong>Register Student Account</strong></h2>
<form action="student_form.php" method="post" >
<table border="1" id="container">
<tr>
<td>First Name</td>
<td>:</td>
<td><input type="text" name="fname" size="30" maxlength="50"/></td>
</tr>
<tr>
<td>Last Name</td>
<td>:</td>
<td><input type="text" name="lname" size="30" maxlength="50"/></td>
</tr>
<tr>
<td>Age</td>
<td>:</td>
<td><input type="text" name="age" size="3" /></td>
</tr>
<tr>
<td >Gender </td>
<td> : </td>
<td> Male
<input type="radio" name="gender" value="Male"/>
Female
<input type="radio" name="gender" value="Female"/></td>
</tr>
<tr>
<td valign="top">Course</td>
<td valign="top"> : </td>
<td> <input type="radio" name="course" value="Bachelor Of Computing"/>Bachelor Of Computing<br/>
<input type="radio" name="course" value="Bachelor Of Science"/>Bachelor Of Science<br/>
<input type="radio" name="course" value="Bachelor Of Software Engineering"/>Bachelor Of Software Engineering<br/>
<input type="radio" name="course" value="Bachelor Of Networking"/>Bachelor Of Networking<br/>
<input type="radio" name="course" value="Bacelor Of IT"/>Bacelor Of IT <br/>
<input type="radio" name="course" value="Bachelor Of Computer Science"/>Bachelor Of Computer Science<br/></td>
</tr>
<tr>
<td>Email Address</td>
<td>:</td>
<td><input type="text" name="email" size="30" maxlength="50"/></td>
</tr>
</table>
<input type="submit" name="register" value="REGISTER"/>
</form><br>
<p><a href="student_form.php" >Home</a></p>
<p><a href="display_data.php">Display Data Information</a>
<p><a href="search.php">To search for Members</a>
</body>
</html>
，這是search_form碼......
<html>
<head>
<?php
//require_once('student_form.php');
if(isset($_POST['s1'])){
$id=$_REQUEST['id'];
$fname=$_POST['fname'];
//connect to the database
include('connection.php');
//-query the database table
$sql=mysql_query("SELECT * FROM members WHERE (FirstName LIKE '". $fname ."%' OR LastName LIKE '". $lname ."%'");
//-run the query against the mysql query function
$result=mysql_query($sql);
if($row=mysql_fetch_array($result)){
$fname=$row['FirstName'];
$lname=$row['LastName'];
/*$email=$row['Email'];
$age =$row['Age'];
$gender=$row['Gender'];
$course = $row['Course'];*/
}
//-display the result of the array
else
{
<?php echo $rows['FirstName']; ?>
<?php echo $rows['LastName']; ?>
}
}
?>
</head>
<body>
<form action="search.php" method="post">
<table>
<tr>
<td><strong>search box</strong></td>
<td><strong>:</strong></td>
<td><input type="text" name="search" value=""size="30"/><input type="submit" name="s1" value="Search"/></td>
</table>
</form>
</body>
</html>
Wowwwww！它的工作..謝謝很多...我真的很感激它..但你可以幫助我只有1件事情，我有一個搜索頁面，代碼在上面給出，例如：用戶輸入名稱或姓氏或電子郵件，然後按搜索，這將顯示所需成員從數據庫中的結果到頁面。 – bleach64
你的sql查詢似乎有錯誤... WHERE子句後面有一個括號，不需要，請嘗試刪除它 – Shades88