Question

我觉得我只需要另外一套眼睛。当然，数据库中有一些东西需要搜索，但是没有显示任何内容。语法或逻辑是否有问题。这是一个文件index.php

    <form action = "index.php" method = "post">
 Search: <input type="text" name="value" placeholder="Is it part of the FWO?"></input>
  <input type=submit name = "search" value="Search">
    </form> 

  <a href="LINKY">New Entry</a>
  <br>
  <p>Search Results</p>
  <hr /> 


  <?php
       error_reporting(E_ALL);
       $title = $_POST['value'];
       echo "You have searched: " .$title;
       echo "<br>";

       $con = mysql_connect("localhost", "user", "pass") or die ('Could not connect, this is the error:  ' . mysql_error());
       mysql_select_db("db") or die ('Sorry could not access database at this time.  This is the error:  ' . mysql_error());   

       $clean = msql_real_escape_string($_GET['value']);
       echo "Another test ". $clean;
       $run = mysql_query("SELECT * FROM db WHERE name = '$clean'") or die(mysql_error());

       if(mysql_num_rows($run) >= 1){
        echo "found entry";
        while($i = mysql_fetch_array($run)){
            echo $i['creator'];
        }

       }

       else {
        echo "No entries found";
       }


       mysql_close($con);
?>



  </body> 
</html>

Answer 1

您的表单正在使用post方法，并且您正在尝试通过$ _GET

获取值

而不是：

$clean = msql_real_escape_string($_GET['value']);

使用此：

$clean = msql_real_escape_string($_POST['value']);

或者

$clean = msql_real_escape_string($title);

Answer 2

要在mysql中搜索，你应该使用LIKE。如果你想搜索字符串中的任何地方，你应该用％封装。例如：

   $run = mysql_query("SELECT * FROM db WHERE name LIKE '%$clean%'") or die(mysql_error());

了解更多信息：http://dev.mysql.com/doc/refman/5.7/en/string-comparison-functions.html

php / mysql搜索功能问题

2 个答案: