如何解決SQL查詢中的單引號錯誤？

Question

我有一個SQL查詢，其中我將英寸轉換爲英尺。

<?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); 
?> 

從我明白我應該單引號，我已經做了

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

雙人和我也試過這個

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

無論是一期工程。我嘗試了其他一些組合，但總是在最後一行中出現錯誤。

我跟着回答這個問題 - How do I escape a single quote in SQL Server?但我「米仍不能確定什麼，我做錯了

Answer 1

你需要逃避雙引號，因此不會結束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 
"; 

但沒有必要在第一時間使用replace，只需使用字符串連接。

$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 
";