2016-10-03 58 views
0

使这个submit.php为学生数据提交到数据库得到错误,当我们从MySQL数据库中的数据如何正确

这里是代码

<html> 
    <head> 
     <title>Submit Student DATA </title> 
     <link rel="stylesheet" type="text/css" href="style.css"> 
    </head> 
    <body> 

     <div id="main"> 
      <h1>Submit Student Data for Verify </h1> 
      <div id="login"> 
       <h2>Student's Form</h2> 
       <hr/> 
       <form action="" method="post"> 
        <label>Student Name :</label> 
        <input type="text" name="stu_name" id="name" required="required" placeholder="Please Enter Name"/><br /><br /> 
        <label>Father Name :</label> 
        <input type="text" name="stu_fathername" id="name" required="required" placeholder="Please Father Name"/><br /><br /> 
        <label>Phone Number :</label> 
        <input type="text" name="stu_phonenumber" id="name" required="required" placeholder="+92000000"/><br /><br /> 
        <label>Address :</label> 
        <input type="text" name="stu_address" id="name" required="required" placeholder="Pakistan, Punjab 0000"/><br /><br /> 
        <label>Course Name :</label> 
        <input type="text" name="stu_course" id="name" required="required" placeholder="Nebosh saftey"/><br /><br /> 
        <label>Certificate Number :</label> 
        <input type="text" name="stu_certificatenumber" id="name" required="required" placeholder="Enter Number"/><br /><br /> 
        <label>Registration Number :</label> 
        <input type="text" name="stu_registrationnumber" id="name" required="required" placeholder="Enter Number"/><br /><br /> 
        <label>Student Email :</label> 
        <input type="email" name="stu_email" id="email" required="required" placeholder="[email protected]"/><br/><br /> 
        <label>Student City :</label> 
        <input type="text" name="stu_city" id="city" required="required" placeholder="Lahore"/><br/><br /> 
        <input type="submit" value=" Submit " name="submit"/><br /> 
       </form> 
      </div> 


     </div> 
     <?php 
     if (isset($_POST["submit"])) { 
      $servername = "localhost"; 
      $username = "root"; 
      $password = "root"; 
      $dbname = "college"; 

// Create connection 
      $conn = new mysqli($servername, $username, $password, $dbname); 
// Check connection 
      if ($conn->connect_error) { 
       die("Connection failed: " . $conn->connect_error); 
      } 

      $sql = "INSERT INTO students (student_name, father_name, phone_number, student_address, student_course, student_certificatenumber, student_reg, student_email, student_city) 
VALUES ('" . $_POST["stu_name"] . "','" . $_POST["stu_fathername"] . "','" . $_POST["stu_phonenumber"] . "','" . $_POST["stu_address"] . "','" . $_POST["stu_course"] . "','" . $_POST["stu_certificatenumber"] . "','" . $_POST["stu_registrationnumber"] . "','" . $_POST["stu_email"] . "','" . $_POST["stu_city"] . "')"; 

      if ($conn->query($sql) === TRUE) { 
       echo "<script type= 'text/javascript'>alert('New Record Inserted Successfully');</script>"; 
      } else { 
       echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error . "');</script>"; 
      } 

      $conn->close(); 
     } 
     ?> 
    </body> 
</html> 

我想要的搜索框,在1个文本box call student注册号码。当我把学生注册人数和点击搜索,然后我从MySQL获得

这里的学生数据搜索代码也许正确或wroing我不无

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<?php 
$hostname_php_result_conn = "localhost"; 
$database_php_result_conn = "college"; 
$username_php_result_conn = "root"; 
$password_php_result_conn = "root"; 
$php_result_conn = mysql_connect($hostname_php_result_conn, $username_php_result_conn, $password_php_result_conn); 
mysql_select_db($database_php_result_conn,$php_result_conn); 

if(isset($_POST['submit'])) { 
    $student_reg = $_POST['stu_registrationnumber']; 

    $sql1 = mysql_query("SELECT students FROM student_reg WHERE = $stu_registrationnumber"); 
    $row1 = mysql_num_rows($sql1); 

    if($row1 == 0) { 
     echo 'Error, ID does not exist'; 
    } else { 
     header('Location: show_result.php?student_reg='.$stu_registrationnumber); 
    } 
} 
?> 

<html> 
<head> 
<title>Search</title> 
</head> 
<body> 

<form action="search.php" method="post"> 
<input name="student_reg" type="text" id="stu_registrationnumber" /> 
<input type="submit" name="submit" value="search" /> 
</form> 
</body> 
</html> 

,这里是

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<?php 
$hostname_php_result_conn = "localhost"; 
$database_php_result_conn = "college"; 
$username_php_result_conn = "root"; 
$password_php_result_conn = "root"; 
$php_result_conn = mysql_connect($hostname_php_result_conn, $username_php_result_conn, $password_php_result_conn); 
mysql_select_db($database_php_result_conn,$php_result_conn); 

$stu_reg = $_GET['stu_registrationnumber']; 

$sql1 = mysql_query("SELECT * FROM students WHERE stu_registrationnumber = $stu_registrationnumber"); 
$row1 = mysql_fetch_array($sql1); 
$student_name = $row1['student_name']; 
$phone_number = $row1['phone_number']; 
$student_address = $row1['student_address']; 
$student_course = $row1['student_course']; 
$student_certificatenumber = $row1['student_certificatenumber']; 
$stu_registrationnumber = $row1['stu_registrationnumber']; 
$student_email = $row1['student_email']; 
$student_city = $row1['student_city']; 
?> 

<html> 
<head> 
<title>Show Results</title> 
</head> 
<body> 

<?php 
echo 'Student Name = '.$student_name.'<br /> 
Father Name = '.$father_name; 
Phone Number = '.$phone_number; 
Address = '.$student_address; 
Course = '.$student_course; 
Certificate Number = '.$student_certificatenumber; 
Registration Number = '.$stu_registrationnumber; 
E-Mail = '.$student_email; 
City = '.$student_city; 
?> 

</body> 
</html> 
show_result.php页

,但我得到的错误,ID不存在,但在数据库中,我们已经登记号码,但得到的错误还是错误

+0

'$ student_reg = $ _POST [ 'student_reg'];'VS'' - $ _POST ['stu_registrationnumber']' –

+0

'$ _POST ['student_reg'];' - >'$ _POST ['stu_registrationnumber'];'**但是** don不使用mysql_ *,清理用户输入,使用预处理语句。 – Jeff

+0

另外,为什么你在一个中使用mysqli_ API,然后在另一个中使用mysql_? –

回答

0

你有一个错误的岗位价值。

变化:

<input type="text" name="stu_registrationnumber" /> 

<input type="text" name="student_reg" /> 
+0

ahaha打了我几秒钟。有趣的是,我们的帖子在格式上几乎相同。 –

+0

@MatthewGoulart哈哈我注意到了。这对他来说更容易继续他的项目 –

+0

我这样做,但仍然得到空白页 –

0

的名称,如果你的学生注册输入为 “stu_registrationnumber”,但你正在寻找Display.php的 “student_reg”。

变化

$student_reg = $_POST['student_reg']; 

$student_reg = $_POST['stu_registrationnumber'] 
+0

我这样做,但仍然在display.php –

+0

空白页在这种情况下需要更多的信息。你有错误吗?错误报告是否打开? 'echo $ _POST ['stu_registrationnumber']'的结果是什么?如果你在工作台这样的工具中运行你的查询,你会得到一个mysql错误吗?我建议你自己做一些故障排除。最简单的方法是将'var_dump'放在关键位置,以确保您没有使用意外的NULL。 –

+0

是的,我也这样做,但仍然空白页面的display.php –

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