PHP解析錯誤：語法錯誤

我有一個mysqli_query這樣的：PHP解析錯誤：語法錯誤

SELECT a.* FROM (SELECT `id` as `id`, `age` as `age` FROM `register` WHERE `age` !="") as a INNER JOIN (SELECT `one` as `f1` FROM `friends` WHERE `two`='".$my_id."' UNION SELECT `two` as `f2` FROM `friends` WHERE `one` = '".$my_id."') as b ON a.id=b.f1

當我在SQL運行這個它沒有給出錯誤並顯示一個成功的查詢，但如果我在瀏覽器中運行它，我得到這個錯誤

(!) Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in C:\wamp\www\functions.php on line 585

請任何人有一個想法，我怎麼能解決這個問題？

這是我的全部代碼

<?php 
    $con = mysqli_connect('localhost','root',''); 

    mysqli_select_db($con, 'qings'); 

    $my_id = '1'; 
    // the line below is my line 585 
    $query = mysqli_query($con, "SELECT a.* FROM (SELECT `id` as `id`, `age` as `age` FROM `register` WHERE `age` !="") as a INNER JOIN (SELECT `one` as `f1` FROM `friends` WHERE `two`='".$my_id."' UNION SELECT `two` as `f2` FROM `friends` WHERE `one` = '".$my_id."') as b ON a.id=b.f1"); 

?>

來源

2016-07-03 KANAYO AUSTIN KANE

請提供一些代碼 – AkshayP

顯示相關的PHP代碼，錯誤不在SQL查詢中 – LeleDumbo

您需要發佈生成錯誤的PHP代碼。如錯誤消息所示，它位於第585行的'C：\ wamp \ www \ functions.php'中。 – mhawke

變化，其中 '年齡'！= 「」到WHERE '年齡'！=」 '

使用

$query = mysqli_query($con, "SELECT a.* FROM (SELECT `id` as `id`, `age` as `age` FROM `register` WHERE `age` !='') as a INNER JOIN (SELECT `one` as `f1` FROM `friends` WHERE `two`='".$my_id."' UNION SELECT `two` as `f2` FROM `friends` WHERE `one` = '".$my_id."') as b ON a.id=b.f1");

來源

2016-07-03 12:25:04

您應該只是逃避這裏的雙引號

來源

2016-07-03 12:05:21 Nick

PHP解析錯誤：語法錯誤

回答

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