2017-01-18 80 views
0

如果找不到結果,我需要一些幫助在搜索框下方的頁面上彈出「無結果」消息。任何援助將不勝感激。當搜索沒有結果時添加「無結果」消息

目前代碼:

<html lang="en"> 
<head> 
<meta charset="utf-8" /> 
</head> 
<body> 
<form action="" method="post"> 
Search: <input type="text" placeholder="Enter Address" name="term" /><br /> 
<br /> 
<button class="ladda-button btn btn-primary" data-style="zoom-in">Submit 
</button> 
</form> 
<?php 
    if (!empty($_REQUEST['term'])) 
    { 
     $term = mysql_real_escape_string($_REQUEST['term']);  

$sql = "SELECT * FROM triadlocations WHERE address LIKE '%".$term."%'"; 
$r_query = mysql_query($sql); 

    while ($row = mysql_fetch_array($r_query)) 
    { 
    echo '<br /><br /> Location ID: ' .$row['locationid']; 
    echo '<br /> Address: ' .$row['address']; 
    echo '<br /> City: '.$row['city']; 
    echo '<br /> State: '.$row['state']; 
    echo '<br /> Zip: '.$row['zip']; 
    } 

} 
?> 
</body> 
</html> 
+0

許多簡單的解決方案存在,如添加初始化爲假的變量,並將其設置在真正的同時,事後檢查,或把所有行陣列中的第一和檢查,如果數組爲空,否則環..等等 –

回答

0

檢查您while循環之前的結果數量。如果沒有結果,打印消息。

if (mysql_num_rows($r_query)) { 
    while ($row = mysql_fetch_array($r_query)){ 
    ... 
    } 
} else { 
    echo "No Results"; 
}