PHP，MySQL验证故障和搜索不起作用？

时间：2012-07-02 12:28:53

标签： php mysql html phpmyadmin

我创建了一个小型注册粘贴表单。一切都工作正常，但如果我输入任何错误的值，如姓名中的数字，年龄的字母，甚至错误的电子邮件格式，那么数据仍保存在数据库中，我无法弄清楚验证问题。 另外一个是搜索选项。每当我在搜索框中输入任何FirstName或lastname时，它应显示数据库中的名称，否则将显示错误消息。 任何1可以建议我该怎么做..请参阅下面的编码。

sticky_form代码...... 

<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.-]+@[\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>

这是search_form代码...... 

<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>

2 个答案:

答案 0 :(得分：1)

$fname $lname $gender $age $email $course之类的变量如果条件为isset($_POST['register'])之后的条件，则会输入if($fname&&$lname&&$gender&&$age&&$email&&$course)。现在，即使您的验证有效，数据仍将输入数据库。因为你已经把条件

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>'; $fname = ""; }

现在，当所有这些变量中只有一个值时，控件将进入该块。必须发生的是，你输入了错误的值，那些正在验证的信息，将显示消息，但是当第一个if块完成时，因为$ _POST变量仍然具有SOME值，无论它们是否无效，第二个if block将被输入和查询将被解雇。

您可以做的是，无论您何时回显错误消息，都会删除相应的变量。像这样的东西：

{{1}}

答案 1 :(得分：0)

关于您的第一个问题，请不要检查$var，检查!empty($var)。

第二个问题：我不确定我明白你要做什么。但是，当您展示某些内容时，不是指$row而不是$rows吗？此外，如果您有多个搜索匹配怎么办？你想要的是：

while($row=mysql_fetch_assoc($result)){
   echo $row['Firstname'], ' ', $row['LastName'];
}

而不是你的if-else结构。

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