如何解決SQL查詢中的單引號錯誤？

Question

我有一個SQL查詢，其中我將英寸轉換為英尺。

<?php  
$query ="
SELECT *
,replace (replace('<feet>'' <inches>"',
                   '<feet>', height / 12),
           '<inches>', height % 12) AS playerHeight
,abbrName
FROM leagueRosters
INNER JOIN leagueTeams AS teamInfo
ON leagueRosters.teamId=teamInfo.teamId

WHERE abbrName LIKE '".$_GET['team']."'
ORDER BY rosterId DESC
";  
$resultRoster = mysqli_query($connect, $query);  
?>

據我了解，我應該將單引號加倍，這已經做了

,replace (replace("<feet>"" <inches>"",
                   "<feet>", height / 12),
           "<inches>", height % 12) AS playerHeight

我也嘗試過

,replace (replace("'<feet>'" "'<inches>'",
                       "'<feet>'", height / 12),
               "'<inches>'", height % 12) AS playerHeight

兩者均無效。 我嘗試了其他幾種組合，但總是至少在一行上出現錯誤。

我遵循了這個問題的答案- 如何在SQL Server中轉義單引號？ 但是我仍然不確定我在做什么錯。

Answer 1

您需要轉義雙引號，以便它不會結束PHP字符串。

$query ="
SELECT *
,replace (replace('<feet>'' <inches>\"',
                   '<feet>', height / 12),
           '<inches>', height % 12) AS playerHeight
,abbrName
FROM leagueRosters
INNER JOIN leagueTeams AS teamInfo
ON leagueRosters.teamId=teamInfo.teamId

WHERE abbrName LIKE '".$_GET['team']."'
ORDER BY rosterId DESC
";

但是首先不需要使用replace ，只需使用字符串連接即可。

$query ="
SELECT *
,CONCAT(FLOOR(height/12), ''' ', height % 12, '\"') AS playerheight
,abbrName
FROM leagueRosters
INNER JOIN leagueTeams AS teamInfo
ON leagueRosters.teamId=teamInfo.teamId

WHERE abbrName LIKE '".$_GET['team']."'
ORDER BY rosterId DESC
";