以下是我创建的PHP文件的代码。
<?php
$db = mysqli_connect("127.0.0.1","root","toor","mylib")
or die('Error connecting to MySQL server.');
?>
<html>
<head>
</head>
<h1>PHP connect to MySQL</h1>
</body>
</html>
<?php
$query = "SELECT * FROM book WHERE bookid IN (SELECT bookid FROM studentbook WHERE studid =
$_POST['stuid'])";
mysqli_query($db, $query) or die('Error querying database.');
$result = mysqli_query($db, $query);
$row = mysqli_fetch_row($result);
foreach($row as $a)
print($a." ");
?>
现在,我希望从HTML登录页面获取studid
并传递它
使用POST
方法从PHP文件中检索与MySQL
数据库中输入的值(studid)对应的数据 - 我的意思是使用mysqli_query()
函数。
但是当我尝试运行上面显示的代码时,我想出了Error
下面的代码:
Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in /opt/lampp/htdocs/connectivity.php on line 13
请指导我以下内容:
1)如何从Login page
获取值并使用PHP file
方法将其传递给post
?
2)如何纠正上述错误?
答案 0 :(得分:1)
希望以下代码必须正常工作,您的错误可能会得到解决,
<?php
$db = mysqli_connect("127.0.0.1","root","toor","mylib")
or die('Error connecting to MySQL server.');
/* get values from form */
$sid = $_REQUEST['stuid'];
/* execute query */
$query = "SELECT * FROM book WHERE bookid IN (SELECT bookid FROM studentbook WHERE studid = '$sid')";
mysqli_query($db, $query) or die('Error querying database.');
$result = mysqli_query($db, $query);
$row = mysqli_fetch_row($result);
foreach($row as $a)
print($a." ");
?>
现在,您登录页面中的数据将存储在$sid
变量中,并将在查询中进行插值。因此,如果您的数据库已连接且一切正常,您可以看到输出存储在$row
数组中。