php - 从数据库中检查和返回值

Question

我有这个小问题:

我想从数据库中检索一些行，它检索正常，但是当我在文本字段中插入一个特定值时，该值不存在于数据库中进行跟踪，它返回一个空/空白页，下面是代码我用过:

<form id="track" name="track" method="post" action="track_now.php"> 
        <h2>Track your shipment Here</h2>

      <p><label> Tracking Reference: 
      <input type="text" id="reference" name="reference" value="" maxlength="40" required="required" /></label></p>


      <div class="button_holder">

        <p>   <input type="submit" id="track" value="Track Now" maxlength="40" required="required" /></label>
      </label></p>

      </div>
</form>

这是 track_now.php

<form id="track" name="track" method="post" action=""> 
        <h2>Your Shipment Result</h2>

            <?php
//error_reporting(0);
$ref = mysql_real_escape_string($_POST['reference']);

// conmnecting to the database
if(isset($ref))
{ 
$db = mysql_connect('localhost', 'admin', "admin") or die(mysql_error("Cannot Connect to Database")); 
mysql_select_db('tracking') or die(mysql_error());

 $sql = "SELECT * FROM order_tracking WHERE ship_ref = '".$ref."' "; 

$rs  = mysql_query($sql);
if($row = mysql_fetch_array($rs)) {

echo '<table width="518" border="1";>'; 

         echo '<tr>';
 echo '<td width="137" style="font-size:12px; padding: 5px;" >Shipment Reference: </td>';
  echo '<td width="365" style="background-color:#fcfcfc; padding: 10px;  font-size:12px;">' . $row['ship_ref'] . "<br />" . '</td>'; 
 echo '</tr>';

     echo '<tr>';
 echo '<td width="137" style="font-size:12px; padding: 5px;" >Shipment Type: </td>';
 echo '<td width="365" style="background-color:#fcfcfc; padding: 10px;  font-size:12px;">' . $row['ship_type'] . "<br />" . '</td>'; 
 echo '</tr>';

}

echo "</table>";
}
else if ($rs != $row) {
print 'Invalid Tracking Number, Please <a href="tracking.php"> click here </a> to try  again' ;
}

mysql_close();
?>    

拜托，我做错了什么？

Answer 1

你的条件可以简化，试试这个:

if(isset($ref)){ 
    $db = mysql_connect('localhost', 'admin', "admin") or die(mysql_error("Cannot Connect to Database")); 
    mysql_select_db('tracking') or die(mysql_error());

    $sql = "SELECT * FROM order_tracking WHERE ship_ref = '".$ref."' "; 
    $rs  = mysql_query($sql);
    if(!rs){
        die(mysql_error());
    }

    if($row = mysql_fetch_array($rs)) {
        echo '<table width="518" border="1";>'; 

        echo '<tr>';
        echo '<td width="137" style="font-size:12px; padding: 5px;" >Shipment Reference: </td>';
        echo '<td width="365" style="background-color:#fcfcfc; padding: 10px;  font-size:12px;">' . $row['ship_ref'] . "<br />" . '</td>'; 
        echo '</tr>';

        echo '<tr>';
        echo '<td width="137" style="font-size:12px; padding: 5px;" >Shipment Type: </td>';
        echo '<td width="365" style="background-color:#fcfcfc; padding: 10px;  font-size:12px;">' . $row['ship_type'] . "<br />" . '</td>'; 
        echo '</tr>';
        echo "</table>";
    }
    mysql_close();
}
else{
    print 'Invalid Tracking Number, Please <a href="tracking.php"> click here </a> to try  again' ;
}

因为如果不满足第一个条件，else if ($rs != $row) 将具有未定义的值。

正如@marco 指出的，您可以在不获取的情况下检查行:

if(mysql_num_rows($rs) > 0){
  //found a row
}