在PhP中搜索MySQL数据库

Question

I am trying to search a MySQL database in PhP using POST , but I get the following errors:

Notice: Undefined variable: ban1 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 266

Notice: Undefined variable: banner1 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 268

Notice: Undefined variable: bantl1 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 270

Notice: Undefined variable: banreason1 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 274

Notice: Undefined variable: ban2 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 276

Notice: Undefined variable: banner2 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 278

Notice: Undefined variable: bantl2 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 280

Notice: Undefined variable: banreason2 in /Applications/XAMPP/xamppfiles/htdocs/bans.php on line 286

Here is my code:

<h1>Punishments
 <div class='col-sm-3 col-md-3 pull-right' style='float:right'>
<div class='container'>
<div class='row'>
<div class='col-sm-3 col-xs-12 col-lg-3'>
  <form class='form-search' action='./bans.php' method='post'>
      <div class='input-group'>
          <input type='text' class='form-control' name='psch' placeholder='Search   Players'>
          <span class='input-group-btn'>
              <button type='submit' name='submitbtn' class='btn btn-search'><span   class='glyphicon glyphicon-search'></span></button>
          </span>
      </div>
  </form>
</div>
</div>
</div>
</div>
</h1>
<br>
<hr>";
$ban1 = "";

    require ("./connect.php");
if (isset($_POST['submitbtn'])) {
    $search = $_POST['psch'];
            $searched = mysql_real_escape_string($search);
    $result = mysql_query('SELECT * FROM bans WHERE banned="$searched" ORDER BY id DESC LIMIT 0,1') or die('Invalid query: ' . mysql_error());
    while ($row = mysql_fetch_assoc($result)) {
        $ban1 = $row['banned'];
        $banner1 = $row['banner'];
        $bantl1 = $row['timeleft'];
        $banreason1 = $row['reason'];
        $banappeal1 = $row['appealed'];
        $banacceptor1 = $row['acceptor'];
        $bantime1 = $row['time'];
        }
    $result = mysql_query('SELECT * FROM bans WHERE banned="$searched" ORDER BY id DESC LIMIT 1,1') or die('Invalid query: ' . mysql_error());
        while ($row = mysql_fetch_assoc($result)) {
            $ban2 = $row['banned'];
        }
    $result = mysql_query('SELECT * FROM bans WHERE banned="$searched" ORDER BY id DESC LIMIT 2,1') or die('Invalid query: ' . mysql_error());
        while ($row = mysql_fetch_assoc($result)) {
            $ban3 = $row['banned'];
        }
    $result = mysql_query('SELECT * FROM bans WHERE banned="$searched" ORDER BY id DESC LIMIT 3,1') or die('Invalid query: ' . mysql_error());
        while ($row = mysql_fetch_assoc($result)) {
            $ban4 = $row['banned'];
        }
    $result = mysql_query('SELECT * FROM bans WHERE banned="$searched" ORDER BY id DESC LIMIT 4,1') or die('Invalid query: ' . mysql_error());
        while ($row = mysql_fetch_assoc($result)) {
            $ban5 = $row['banned'];
        }
    echo "
        <div class='jumbotron'>
        <h1></h1>
            <p>
   <b>-----------------------------------------------</b>
   <br>
   <b>Banned:</b>$ban1
  <br>
    <b>Banner:</b>$banner1
<br>
<b>Time Left:</b>$bantl1
<br>
<b>Reason:</b>$banreason1
<br>
<b>-----------------------------------------------</b>
<br>
<b>Banned:</b>$ban2
<br>
<b>Banner:</b>$banner2
<br>
<b>Time Left:</b>$bantl2
<br>
<b>Reason:</b>$banreason2
<br>
<b>-----------------------------------------------</b>
</p>
            <p></p>
            </div>
        ";

    }
    else

I have tried declaring ban1 , banner1 , and all the other variables outside of the while statement, yet I still get those errors. Any ideas?

Answer 1

You have to concatenate php variables with html like this

echo "
    <div class='jumbotron'>
    <h1></h1>
    <p>
    <b>-----------------------------------------------</b>
    <br>
    <b>Banned:</b>" . $ban1 .
    "<br>
    <b>Banner:</b>" . $banner1;

Answer 2

From PHP docs, you can see that

E_NOTICE level error is issued in case of working with uninitialized variables, however not in the case of appending elements to the uninitialized array. isset() language construct can be used to detect if a variable has been already initialized. You can find more info here

So you have to initialize the variables inside your while loop before you use it to get rid of the notice.

Answer 3

Are you sure that

if (isset($_POST['submitbtn'])) { 

is true? you might want to add an else statement to confirm. or do you get the same error when you know it is false?