檢查，並從數據庫

Question

返回值我有這個小問題：

我想從數據庫中檢索某些行，它檢索很好，但是當我插入的文本字段中的特定值，它是不存在的數據庫來跟蹤，它返回一個空/空白頁，下面是我使用的代碼：

<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> 

，這是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(); 
?>  

拜託，我是什麼錯在這裏做什麼？

Answer 1

你的情況可以簡化，試試這個方法：

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' ; 
} 

因爲else if ($rs != $row)如果第一個條件不滿足，就會有明確的值。

由於@marco指出的那樣，你可以檢查行，而不取：

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