我一遍又一遍,但仍然无法解决为什么服务器告诉我它没有期待';'在第19行。以下是:
( '$ _ POST [modulecode]', '$ _ POST [学校]', '$ _ POST [日期]')“;
<?php
$link=mysqli_connect("bla","bla","bla","bla");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else
{
echo "Connection Succesfull";
}
mysqli_query($link,"INSERT INTO student (tester, testertwo, testerthree)
VALUES
('$_POST[modulecode]','$_POST[school]','$_POST[date]')";
if (!mysqli_query($link,$sql))
{
die('Error: ' . mysqli_error($link));
}
echo "1 record added";
mysqli_close($link);
?>
答案 0 :(得分:2)
您错过了)
的右括号mysqli_query()
:
mysqli_query($link,"INSERT INTO student (tester, testertwo, testerthree)
VALUES
('$_POST[modulecode]','$_POST[school]','$_POST[date]')");
答案 1 :(得分:0)
你错过了那里的右括号
mysqli_query($link,"INSERT INTO student (tester, testertwo, testerthree) VALUES ('$_POST[modulecode]','$_POST[school]','$_POST[date]')");
一个建议总是先用简单的php回复你的查询...然后把它放到mysqli_query
函数中..
答案 2 :(得分:0)
我喜欢用这种方式写它的部分原因(虽然我也会逃避任何和所有输入,无论我认为它来自哪里)......
$query = "
INSERT INTO student
(tester
,testertwo
,testerthree
) VALUES
('$_POST[modulecode]'
,'$_POST[school]'
,'$_POST[date]')
";
mysqli_query($link,$query);
答案 3 :(得分:0)
除了@Sirko所说的,我认为用括号之类的方式包装变量(包括字符串内部)是个好主意:
mysqli_query($link,"INSERT INTO student (tester, testertwo, testerthree) VALUES ('{$_POST[modulecode]}','{$_POST[school]}','{$_POST[date]}')");