phpmyadmin - count():参数必须是数组或实现Countable的对象

我已经上传了备份到一个表，打开表我看到这个:

Warning in ./libraries/sql.lib.php#601
count(): Parameter must be an array or an object that implements Countable

Backtrace

./libraries/sql.lib.php#2038: PMA_isRememberSortingOrder(array)
./libraries/sql.lib.php#1984: PMA_executeQueryAndGetQueryResponse(
array,
boolean true,
string 'alternativegirls',
string 'tgp_photo',
NULL,
NULL,
NULL,
NULL,
NULL,
NULL,
string '',
string './themes/pmahomme/img/',
NULL,
NULL,
NULL,
string 'SELECT * FROM `tgp_photo`',
NULL,
NULL,
)
./sql.php#216: PMA_executeQueryAndSendQueryResponse(
array,
boolean true,
string 'alternativegirls',
string 'tgp_photo',
NULL,
NULL,
NULL,
NULL,
NULL,
NULL,
string '',
string './themes/pmahomme/img/',
NULL,
NULL,
NULL,
string 'SELECT * FROM `tgp_photo`',
NULL,
NULL,
)
./index.php#53: include(./sql.php)

在phpMyAdmin……

PHP是7.2，服务器是Ubuntu 16.04，昨天安装的。

我看到一些人在他们的代码中有这个错误，但我没有发现任何人在phpMyAdmin中收到它…

我该怎么办?这是我的错误吗?phpmyadmin错误?等待更新?回到PHP 7.1?

当前回答

phpmyadmin 4.7.4应该有“修复了与PHP 7.2的几个兼容性问题”

您可能有一个较旧版本的phpmyadmin。

https://www.phpmyadmin.net/news/2017/8/24/phpmyadmin-474-released/

2017-12-29 22:59:56

其他回答

来看看:https://devanswers.co/problem-php-7-2-phpmyadmin-warning-in-librariessql-count/

首先，在编辑之前备份sql.lib.php。在nano中编辑sql.lib.php。

$ sudo nano /usr/share/phpmyadmin/libraries/sql.lib.php

按CTRL + W搜索(count($analyzed_sql_results['select_expr'] == 1)

将其替换为((count($analyzed_sql_results['select_expr']) == 1)

保存文件并退出。(按CTRL + X，按Y，然后按ENTER)

2019-05-06 13:20:14

Works on UBUNTU 18.04 


Edit file: '/usr/share/phpmyadmin/libraries/sql.lib.php'
Replace: (count($analyzed_sql_results['select_expr'] == 1)
With:  ((count($analyzed_sql_results['select_expr']) == 1)

Restart the server
sudo service apache2 restart

2018-11-05 12:22:50

升级到phpMyAdmin 4.8.3。这解决了PHP 7.2的兼容性问题

2018-08-25 17:42:33

替换功能:

function PMA_isRememberSortingOrder($analyzed_sql_results) {
return $GLOBALS['cfg']['RememberSorting']
    &&!(
        $analyzed_sql_results['is_count']
        || $analyzed_sql_results['is_export']
        || $analyzed_sql_results['is_func']
        || $analyzed_sql_results['is_analyse']
    )&&
    $analyzed_sql_results['select_from']&&
    (
        empty($analyzed_sql_results['select_expr'])||
        count($analyzed_sql_results['select_expr'])==1&&
        $analyzed_sql_results['select_expr'][0] == '*'
    )
    && count($analyzed_sql_results['select_tables']) == 1;
}

2019-05-13 21:05:03

在Ubuntu 18.04上，使用MariaDb和Nginx，我通过更新文件/usr/share/phpmyadmin/libraries/sql.lib.php解决了这个问题，如下:

|| (count($analyzed_sql_results['select_expr']) == 1

@Nguyen提到的答案抛出了一个500错误，说:

FastCGI sent in stderr: "PHP message: PHP Parse error:  syntax error, unexpected ')', expecting ';' in /usr/share/phpmyadmin/libraries/sql.lib.php on line 614"

2018-09-10 06:41:50

phpmyadmin - count():参数必须是数组或实现Countable的对象

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