php echo如果查詢爲空

Question

我在查詢數據庫，但是當結果爲空時，我想輸出一個表格行顯示「什麼都不顯示」，但if似乎總是返回true。

這裏是我的代碼...

Answer 1

使用mysqli_num_rows()檢查是否有任何結果：

$conn = mysqli_connect($host, $user, $password, $database); 
    $priorityincidentsQ = mysqli_query($conn, "SELECT * FROM applications WHERE pi >= ('2') "); 
    if (mysqli_num_rows($priorityincidentsQ) > 0){ 
     while ($priorityincidentsR = mysqli_fetch_object($priorityincidentsQ)) { 
      echo "<tr><td class=\"closedcallscell\"><b>$priorityincidentsR->application_friendly_name</b></td>"; 
      echo "<td class=\"closedcallscell table_row_small\"><center>$priorityincidentsR->pi</center></td></tr>"; 
     } 
    }else{  
     echo "<tr><td class=\"closedcallscell centered\"><b>Nothing to display</b></td></tr>"; 
    } 

是的，最好使用mysqli_*功能，而不是mysql_*。

Answer 2

仍然沒有想出爲什麼這樣做對我來說不會這樣，我直接在SQL工作臺中試着查詢，一切看起來應該如何，我最終解決了這個問題。

<!-- priority incidents--> 
<?php 

$priorityincidentsQ   = mysql_query("SELECT * FROM applications WHERE pi >= ('1') "); 
while($priorityincidentsR = mysql_fetch_object($priorityincidentsQ)) 
    { 
    echo "<tr><td class=\"closedcallscell\"><b><a href=\"".DIR."?p=$priorityincidentsR->pageID\">$priorityincidentsR->application_friendly_name</a></b></td>"; 
    echo "<td class=\"closedcallscell table_row_small\"><center>$priorityincidentsR->pi</center></td></tr>"; 
    } 

?> 

<!-- if no incidents--> 
<?php 

$incidentNumberofRowsQ  = mysql_query("SELECT COUNT(*)numberofrows FROM applications WHERE pi >= ('1') "); 
while($incidentNumberofRowsR = mysql_fetch_object($incidentNumberofRowsQ)) 
    { 

     if ($incidentNumberofRowsR->numberofrows == '0') 
     { 
       echo "<tr><td class=\"closedcallscell centered\"><b>Currently no priority incidents</b></td>"; 
     } 
    } 

?> 

看起來似乎是一個相當愚蠢的方式去做，但至少它的作品。感謝所有的幫助。 :)