php搜索功能不起作用

时间:2017-03-20 12:06:26

标签: php mysqli

我正在尝试"按名称按钮搜索用户"搜索用户的数据库,显示"结果错误"无法获取用户详细信息。我不知道哪个部分出了问题。请帮我。我们将不胜感激。这是我的代码。

manageruser.php

<?php 
include("include/config.php");

   $name = "";
   $username = "";
   $password = "";
   $ic = "";
   $contact = "";
   $email = "";
   $nationality = "";
   $program = "";
   $firstintake = "";

   function getPosts()
   {
       $posts = array();
       $posts[0] = $_POST['name'];
       $posts[1] = $_POST['username'];
       $posts[2] = $_POST['password'];
       $posts[3] = $_POST['ic'];
       $posts[4] = $_POST['contact'];
       $posts[5] = $_POST['email'];
       $posts[6] = $_POST['nationality'];
       $posts[7] = $_POST['program'];
       $posts[8] = $_POST['firstintake'];
       return $posts;
   }

// Search
   if(isset($_POST['search']))
   {
       $data = getPosts();

       $search_Query = "SELECT * FROM user WHERE u_name = $data[0]";

       $search_Result = mysqli_query($link, $search_Query);

       if($search_Result)
       {
          if(mysqli_num_rows($search_Result))
          {
            while($row = mysqli_fetch_array($search_Result))
            {
                $name = $row['u_name'];
                $username = $row['u_unm'];
                $password = $row['u_pwd'];
                $ic = $row['u_ic'];
                $contact = $row['u_contact'];
                $email = $row['u_email'];
                $nationality = $row['u_national'];
                $program = $row['u_program'];
                $firstintake = $row['u_fintake'];
            }
          }
          else
          {
            echo "No Data For This Name";
          }
       }
       else
       {
        echo "Result Error";
       }
   }

?>

<fieldset>
<legend>Manage User</legend>
<form name="ManForm" method="post" action="manageuser.php">
<table>
    <tr>
        <td>Name:</td>
        <td><input id="name" name="name" type="text" class="input" pattern="[A-Z\s]+" 
        title="Please enter capital letters" value="<?php echo $name; ?>">
        <span>(Full name) *must capital letters</span></td>
    </tr>

    <tr>
        <td>Username:</td>
        <td><input id="username" name="username" type="text" class="input" value="<?php echo $username; ?>"></td>
    </tr>

    <tr>
        <td>Password:</td>
        <td><input id="password" name="password" type="password" class="input" value="<?php echo $password; ?>"></td>
    </tr>

    <tr>
        <td>Identity card /Passport number:</td>
        <td><input id="ic" name="ic" type="text" class="input" value="<?php echo $ic; ?>"></td>
    </tr>

    <tr>
        <td>Contact number:</td>
        <td><input id="contact" name="contact" type="text" class="input" value="<?php echo $contact; ?>"></td>
    </tr>

    <tr>
        <td>Email:</td>
        <td><input id="email" name="email" type="text" class="input" value="<?php echo $email; ?>"></td>
    </tr>

    <tr>
        <td>Nationality:</td>
        <td><input id="nationality" name="nationality" type="text" class="input" value="<?php echo $nationality; ?>"></td>
    </tr>

    <tr>
        <td>Program:</td>
        <td><input id="program" name="program" type="text" class="input" value="<?php echo $program; ?>"></td>
    </tr>

    <tr>
        <td>First intake:</td>
        <td><input id="firstintake" name="firstintake" type="text" class="input" value="<?php echo $firstintake; ?>"></td>
    </tr>
</table>

    <div>
        <input type="submit" name="search" value="  Search User By Name">
        <input type="submit" name="update" value="  Update User Details">
        <input type="submit" name="delete" value="  Delete User ">
    </div>

</form>
</fieldset>

的config.php

<?php

    $link= mysqli_connect("localhost","root","","course_registration_system");

?>

2 个答案:

答案 0 :(得分:0)

这里的问题是你的查询方式错误。
要运行查询,您应该使用预备声明

理解简单地在变量周围添加引号是不够的非常重要,并且最终会导致无数问题,从语法错误到SQL注入。另一方面,由于预处理语句的本质,它是一种防弹解决方案,无法通过数据变量引入任何问题。

因此,对于您运行的每个查询,如果要使用至少一个变量,则必须用占位符替换它,然后准备查询,然后执行它,分别传递变量。

首先,您必须更改查询,添加占位符代替变量。假设您的查询将变为这样

"SELECT * FROM user WHERE u_name = ?"

然后你必须准备它,绑定变量,然后执行

所以你应该这样做

   $stmt = $link->prepare("SELECT * FROM user WHERE u_name = ?");
   $stmt->ind_param("s", $data[0]);
   $stmt->execute();
   $result = $stmt->get_result();
   while($row = mysqli_fetch_array($result))
   {

答案 1 :(得分:-2)

在查询中使用搜索字符串周围的引号,或者尝试使用 -

$ search_Query =“SELECT * FROM user WHERE u_name ='”。$ data [0]。“'”;

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