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这个主题已经发布了好几次，但是这里发布的解决方案似乎没有帮助。我已经检查了分号、空格和查询，但是每当我调用 subjaccess.php 时，我仍然会收到关于 php 的解析错误。

Parse error: syntax error, unexpected T_STRING in /home/a8367888/public_html/subjaccessrequest.php on line 14

这是我的php代码：

<?php
$mysql_hostname = "localhost";
$mysql_database = "database";
$mysql_user = "username";
$mysql_password = "password";

//$pass = $_POST['email'];
$pass = "blahblach@gmail.com";

$bd = mysql_connect($mysql_hostname, $mysql_user, $mysql_password)
or die("Opps something went wrong");
mysql_select_db($mysql_database, $bd) or die("Oops something went wrong");

$query =  "SELECT (a.acode) as "access_id", (s.s_first_name) as "fname",      (s.s_middle_initial) as "mname", (s.s_last_name) as "lname", (a.s_email) as "email", (a.req_subj) as "subjcode"
FROM access a, student s
WHERE a.s_email = s.s_email
AND a.t_email = '$pass' 
AND a.access_status = "pending";

$result = mysql_query($query) or die('MySql Error' . mysql_error());

while ($row = mysql_fetch_array($result))
{
 $res[]=array('aid'=> $row['access_id'],'fname'=> $row['fname'],'mname'=> $row['mname'], 'lname'=> $row['lname'], 'email'=> $row['email'], 'subj'=> $row['subjcode']);
}

echo '$res';

$response['request'] = $res;
print(json_encode($response));
?>

希望得到您的回应。

4

3 回答 3

2

改变这个

$query =  "SELECT (a.acode) as "access_id", (s.s_first_name) as "fname",      (s.s_middle_initial) as "mname", (s.s_last_name) as "lname", (a.s_email) as "email", (a.req_subj) as "subjcode"

至

$query =  "SELECT (a.acode) as 'access_id', (s.s_first_name) as 'fname',      (s.s_middle_initial) as 'mname', (s.s_last_name) as 'lname', (a.s_email) as 'email', (a.req_subj) as 'subjcode'
于 2013-05-03T08:50:05.410 回答
1

您的字符串中有双引号，尽管它被". 这会破坏您的 PHP 代码。但无论如何你都不需要它们。你也应该使用 aJOIN而不是FROM a, b. 并且查询中的所有标识符都应该用 backtics ` 引用。最终查询应如下所示：

$query =  "
SELECT 
  `a`.`acode` AS `access_id`,
  `s`.`s_first_name` AS `fname`,
  `s`.`s_middle_initial` AS `mname`,
  `s`.`s_last_name` AS `lname`,
  `a`.`s_email` AS `email`,
  `a`.`req_subj` AS `subjcode`
FROM `access` `a`
JOIN `students` `s` 
  ON `a`.`s_email` = `s.s_email`
 AND `a`.`t_email` = '$pass' 
 AND `a`.`access_status` = 'pending'
";
于 2013-05-03T08:49:07.357 回答
0

尝试这个

$query =  "SELECT (a.acode) as " . access_id . ", (s.s_first_name) as " . fname . ",      (s.s_middle_initial) as " . mname . ", (s.s_last_name) as " . lname . ", (a.s_email) as "email", (a.req_subj) as " . subjcode
于 2013-05-03T08:56:18.783 回答