在PHP MySQL搜索中找不到結果時顯示消息

Question

我有一個PHP搜索腳本，用於查詢MySQL數據庫。 目前，當沒有顯示結果時，腳本顯示和錯誤。 如果沒有返回任何內容，如何顯示“未找到結果”這樣的消息？

我的PHP腳本是：

<?php

mysql_connect("localhost","username","password");
mysql_select_db("database");

if(!empty($_GET['q'])){
$query=mysql_real_escape_string(trim($_GET['q']));
$searchSQL="SELECT * FROM links WHERE `title` LIKE '%{$query}%'  LIMIT 8";
$searchResult=mysql_query($searchSQL);

while ($row=mysql_fetch_assoc($searchResult)){
    $results[]="<div class='webresult'><div class='title'><a href='{$row['url']}'>{$row['title']}</a></div><div class='desc'>{$row['description']}</div><div class='url'>{$row['url']}</div></div>";
}

echo implode($results);
}

?>

Answer 1

創建一個MYSQL連接並粘貼以下代碼

$sql="SELECT * FROM tablename WHERE columnname LIKE your variable or constant ";
$result=mysql_query($sql);

$count=mysql_num_rows($result);

if($count>=1){if result was found}

else {if result was not found}

?>

Answer 2

if (empty($results)) { 
    echo 'No results found'; 
} else {
    echo implode($results);
}

Answer 3

您可以計算數組中元素的數量，並繼續進行內爆或顯示您提到的消息。

<?php
     if(count($results) > 0){
         echo implode($results);
     }
     else {
         echo "No results were found.";
     }
?>

你也真的不應該使用mysql_ *函數。 使用改進版本（mysqli_ *）或PDO。

Answer 4

請嘗試以下方法：

<?php

mysql_connect("localhost","username","password");
mysql_select_db("database");

if(!empty($_GET['q'])){
$query=mysql_real_escape_string(trim($_GET['q']));
$searchSQL="SELECT * FROM links WHERE `title` LIKE '%{$query}%'  LIMIT 8";
$searchResult=mysql_query($searchSQL);

if(mysql_num_rows($searchResult) <= 0)
{
    echo "No results";
} else {

    while ($row=mysql_fetch_assoc($searchResult)){
        $results[]="<div class='webresult'><div class='title'><a href='{$row['url']}'>{$row['title']}</a></div><div class='desc'>{$row['description']}</div><div class='url'>{$row['url']}</div></div>";
    }

    echo implode($results);
} 
}

?>

另外請使用MySQLi或PDO，因為它更安全，更好用，有些信息可以在下面找到。 我個人更喜歡MySQLi，但PDO中的預處理語句非常好，每次查詢時都會保存一些代碼行;）

MySQLi和PHP

PDO和PHP

Answer 5

<?php

mysql_connect("localhost","username","password");
mysql_select_db("database");

if(!empty($_GET['q'])){
    $query          =   mysql_real_escape_string(trim($_GET['q']));
    $searchSQL      =   "SELECT * FROM links WHERE `title` LIKE '%{$query}%'  LIMIT 8";
    $searchResult   =   mysql_query($searchSQL);

    // the query was run successfully 
    // and it returned at least a result
    if(is_resource($searchResult) && mysql_num_rows($result) > 0){
        while ($row=mysql_fetch_assoc($searchResult)){
            $results[]="<div class='webresult'><div class='title'><a href='{$row['url']}'>{$row['title']}</a></div><div class='desc'>{$row['description']}</div><div class='url'>{$row['url']}</div></div>";
        }

        echo implode($results);
    } else{
        echo 'No Results were found';
    }
}
?>

Answer 6

在while語句中如何使用簡單的計數器：

$i = 0;

while (condition) {
   $i++;
   do stuff;
}

if ($i ==0) {echo 'No results found';}

Answer 7

您還可以使用函數mysql_num_rows ，它將告訴您查詢返回的行數。

$rows = mysql_num_rows($searchResult);
if($rows <= 0){
    /* Code if there are no rows */
}
else{
    /* At least one row has been found */
}

Answer 8

if (mysql_num_rows($searchResult) == 0) {
   echo "some error message"; 
} else { 
  ... process data
}