我有这个陈述
$date = date('Y');
$userid = $userid_string;
$sql="UPDATE fgusers3 SET user_login = ('$date') WHERE username = '$userid'";
这有效,
$sql="UPDATE fgusers3 SET user_login = ('$date') WHERE username = 'ATOMICCOCKROACH'";
但不是我传递用户名
我做了
echo " ".$userid ."<br/> ";
echo " ".$userid_string ."<br/> ";
echo " ".$date ." ";
并正确传递值。
提前谢谢
TABLE SCHEMA
1 id_user int(11) No None AUTO_INCREMENT Change Change
2 name varchar(128) latin1_swedish_ci No None
3 email varchar(64) latin1_swedish_ci No None
4 phone_number varchar(16) latin1_swedish_ci No
5 username varchar(16) latin1_swedish_ci No
6 password varchar(32) latin1_swedish_ci No
7 confirmcode varchar(32) latin1_swedish_ci Yes
8 user_image text latin1_swedish_ci No None
9 user_favouritesong varchar(32) latin1_swedish_ci No
10 user_location tinytext latin1_swedish_ci No
11 user_desc varchar(32) latin1_swedish_ci No 12 user_msg text latin1_swedish_ci No None
13 user_login text latin1_swedish_ci No None
PERHAPS它的功能这是FILE1.php
<?php
function addLoginDate($userid_string)
{
$con=mysqli_connect("XXX","XXX","XXX","XXX");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$date = date('D');
$userid = $userid_string;
echo " ".$userid ."<br/> ";
echo " ".$userid_string ."<br/> ";
echo " ".$date ." ";
$sql="UPDATE fgusers3 SET user_login = '{$date}' WHERE username = '{$userid}'";
if (!mysqli_query($con,$sql)) {
die('Error: ' . mysqli_error($con));
}
echo "<div> 1 record updated </div>";
mysqli_close($con);
}
?>
PRINTS
AtomicCockroach AtomicCockroach 太阳 1记录更新
这是如何被称为这是FILE2.php
<?PHP $usernamekey = "AtomicCockroach";
echo addLoginDate(" ".$usernamekey ." "); ?>
答案 0 :(得分:0)
当我注入这样的变量时它适用于我(假设我有一个记录&#39; ATOMICCOCKROACH&#39;作为用户名:
$date = date('Y');
//$userid = $userid_string;
$userid = "ATOMICCOCKROACH";
$sql="UPDATE fgusers3 SET user_login = '{$date}' WHERE username = '{$userid}'";
答案 1 :(得分:-1)
@RoyalBg得到它
建议我回复sql的权利
这就是我得到的UPDATE fgusers3 SET user_login ='2014'WHERE username ='AtomicCockroach'
似乎用户名中的空格我将尝试
stackoverflow.com/a/2109339/1309575 -
YUP已解决
必须使用
从$ userid中删除空格 $string = str_replace(' ', '', $string);