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

Ubuntu 18.10(2018年12月)

第613、614、615行，替换为:

        || count($analyzed_sql_results['select_expr']) == 1
            && ($analyzed_sql_results['select_expr'][0] == '*'))
    && count($analyzed_sql_results['select_tables']) == 1;

2018-12-18 11:46:23

其他回答

嗨，下面绝对解决了我同样的问题(导入/导出等):

修复Phpmyadmin [plugin_interface.lib.php] + Php7.2 + Ubuntu 16.04的错误

所以…在ubuntu 18.04下，mysql, php7.2: 终端:

sudo gedit /usr/share/phpmyadmin/libraries/plugin_interface.lib.php

找到下面的行(ctrl+f):

if ($options != null && count($options) > 0) {

我是在551行

并更改如下:

if ($options != null && count((array)$options) > 0) {

Ctrl +s保存更改

在终端:ctrl+c for get back prompt…

然后:sudo systemctl restart apache2

“我认为在新的php版本。它不能对unarray类型使用count()或sizeof()。强制参数数组是解决这个错误的简单方法，…”

感谢原作者解决问题!我试着分享它!

2018-10-14 19:53:13

使用以下命令编辑/usr/share/phpmyadmin/libraries/sql.lib.php文件:

sudo nano +613 /usr/share/phpmyadmin/libraries/sql.lib.php

在第613行，count函数的计算结果总是为true，因为$analyzed_sql_results['select_expr']后没有右括号。进行下面的替换可以解决这个问题，然后需要删除第614行上的最后一个右括号，因为它现在是一个额外的括号。

替换:

((empty($analyzed_sql_results['select_expr']))
    || (count($analyzed_sql_results['select_expr'] == 1)
        && ($analyzed_sql_results['select_expr'][0] == '*')))

((empty($analyzed_sql_results['select_expr']))
    || (count($analyzed_sql_results['select_expr']) == 1)
        && ($analyzed_sql_results['select_expr'][0] == '*'))

重启服务器apache:

sudo service apache2 restart

2018-03-26 03:59:25

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

我发现这个PHP 7.2 count() -语法错误在sql.lib.php

这完美地工作在我的配置:

Debian 9, 
PHP 7.2.3-1+0~20180306120016.19+stretch~1.gbp81bf3b (cli) (built: Mar  6 2018 12:00:19) ( NTS )

打开/usr/share/phpmyadmin/libraries/sql.lib.php

更改行:移动==前面的括号

| | ((count ($ analyzed_sql_results [' select_expr ']) = = 1) & & ($ analyzed_sql_results [' select_expr '] [0 ] == '*')))

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;
 }

2018-03-24 10:28:27

编辑/usr/share/phpmyadmin/libraries/sql.lib.php文件 (备份)

"|| (count($analyzed_sql_results['select_expr'] == 1) 
&&($analyzed_sql_results['select_expr'][0] == '*'))) 
&& count($analyzed_sql_results['select_tables']) == 1;"

"|| (count($analyzed_sql_results['select_expr']) == 1) 
&& ($analyzed_sql_results['select_expr'][0] == '*') 
&& (count($analyzed_sql_results['select_tables']) == 1));"

2018-05-17 09:29:35

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

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