尝试将config.php中的函数“覆盖”包含到其他php页面中。
Index.php:
<?php include("./config/config.php");
over_lay(); ?>
//// html code ////
错误是:
Fatal error: Call to a member function connect() on null in ...\config\config.php on line 182
Config.php的一部分:
include("./config/mysql.php");
function over_
我在一个本地服务器(WAMP)的网站上工作,php和数据库工作得很好。但是当我把网站上传到主机服务器时。并尝试登录,则数据库连接将不起作用。当我使用像mysqli::mysqli()这样的DB函数时,它会给我警告。
整个错误如下:
Warning: mysqli::mysqli() [mysqli.mysqli]: (HY000/2002): No connection could be made
because the target machine actively refused it.
in D:\Hosting\XXXXX\html\raw\database\connect.php
当我们在宿主服务器中尝试这段代码时,它的工作非常完美,但是我们将代码移到VPS服务器godaddy中,它会产生错误。
Fatal error: Call to undefined function mysqli_init() in
/public_html/system/database/drivers/mysqli/mysqli_driver.php on line 135
A PHP Error was encountered
Severity: Warning
Message: Cannot modify header information - headers already
哪种格式是正确的?我还不知道具体如何缩进PHP代码。我觉得第二组是正确的,但我还不知道。
集1
$connection = mysqli_connect('localhost', 'ijdbuser', 'ijdbpw');
if (!$connection)
{
$error = 'Unable to connect to the database server.';
include 'error.html.php';
exit();
}
if (!mysqli_set_charset
我正在尝试内部连接WRITES表和BOOK表。以便在页面task8.htm提交时接收authorID。提交之后,您现在看到的task8.php应该为该authorID编写的所有图书提取bookID、booktitle、ISBN字段。我现在在内部连接表格时遇到了麻烦。 我尝试过使用SELECT bookID,booktitle,quantity,name,region FROM BOOK内部连接BOOK.bookID = WRITES.bookID ORDER BY booktitle,quantity,name;的表。 <?php $conn = mysqli_connect(
当我从手推车中删除我的项目时,此错误显示;
Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\FinalsActivity1\delete.php on line 5
Warning: mysqli_error() expects parameter 1 to be mysqli, string given in C:\xampp\htdocs\FinalsActivity1\delete.php on line 5
**代码
<?php
requ
我正在PHP和mysqli中创建一个类似的不喜欢系统;在uid的帮助下,我还通过加入关注者和照片表来获取用户追随者的帖子。现在我希望用户喜欢追随者的照片,但我无法获得图片的个人id。这是我获取用户追随者图片的代码:
$query = "SELECT photos.image_url,photos.email,photos.username,photos.uid,photos.id FROM photos join followers on photos.uid = followers.user_id where followers.uid = '$id' ORDER B
每当我访问我的网站地址/admin时,它就会出现
Warning: mysqli::mysqli() [mysqli.mysqli]: (28000/1045): Access denied for user 'hotstepp_ocart25'@'localhost' (using password: YES) in /home/hotstepp/public_html/system/library/db/mysqli.php on line 7
Warning: mysql_error() expects parameter 1 to be resourc
这将从以下数据库中提取回退值:
<select name="PACKAGE_ID" id="PACKAGE_ID" ng-model="FormData.Phases" class="form-control" required>
<?php
$result=mysqli_query($conn, "select * from unifi WHERE STATUS!='DELETE' ORDER BY PACKAGE_NAME ASC");
while ($row=mysqli
我的代码中有这些错误,我不知道为什么
误差
1. Undefined variable: con in /opt/local/apache2/htdocs/SE/index.php on line 28
2. mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in /opt/local/apache2/htdocs/SE/index.php on line 28
3. Undefined variable: con in /opt/local/apache2/htdocs/SE/index.ph
我是一个运行ubuntu18.04和Apache2、Php (7.3)和Mysql的新手。
当我跑的时候
sudo php -r 'phpinfo();' | grep -i mysqli
我得到的结果是:
/etc/php/7.3/cli/conf.d/20-mysqli.ini,
mysqli
MysqlI Support => enabled
mysqli.allow_local_infile => Off => Off
mysqli.allow_persistent => On => On
m