mysql腳本。 SET請求不起作用

Question

我有一個看起來很簡單的問題，但由於某種原因我無法修復它。也許你可以一眼就解決它。 我有以下腳本。它調用沒有錯誤，但不能正常工作。我將SET請求分爲兩個以更好地解釋問題。我還添加了一些評論，以更好地看。重點是，兩個類似的請求之一不工作...

include_once ("../php/db_connects.php"); // connect to the database 
$query = mysql_query ('SELECT * FROM articles WHERE status = "1"'); //pick all the lines with status (enum) = 1; 
$row = mysql_fetch_array ($query); // general staff 
do { 
$pluses = $row ['pluses']; 
$minuses = $row['minuses']; 
$link = $row['link']; 

$count = $pluses-$minuses; // checking if article has more positive votes than negaitve 
if($count > 0){ 
mysql_query ('UPDATE articles SET rating = 100 WHERE link = "$link"', $db_conx); 
//the first request works (rating is Int) 
mysql_query ('UPDATE articles SET status = "2" WHERE link = "$link"', $db_conx); 
//the second request doesnt show any mistakes, but it doesn't change status on "2". status is Enum. (ps. there is value of "2" in the database) 
} 
echo "link = " . $link . " count = " . $count; 
}while ($row = mysql_fetch_array ($query)); 

謝謝你的時間。

Answer 1

我認爲你必須改變這一點：

mysql_query ('UPDATE articles SET rating = 100 WHERE link = "$link"', $db_conx); 
//the first request works (rating is Int) 
mysql_query ('UPDATE articles SET status = "2" WHERE link = "$link"', $db_conx); 

這樣：

mysql_query ("UPDATE articles SET rating = 100 WHERE link = '$link'", $db_conx); 
//the first request works (rating is Int) 
mysql_query ("UPDATE articles SET status = '2' WHERE link = '$link'", $db_conx); 

的原因是，PHP不會取代單引號中的字符串變量。