php-7

How do I change the default PHP base-version on shell?

馋奶兔 提交于 2021-02-02 09:09:43
问题 I have Centos 7 and Virtualmin installed, with the tipycal php-fpm 5.4, 7.0, 7.1 that you can choice between the versions you prefer on every virtualhost via Virtualmin control panel, and everything works well. But when I access to the server via SSH and check php -v I get this: PHP 5.4.16 (cli) (built: Oct 30 2018 19:30:51) Copyright (c) 1997-2013 The PHP Group Zend Engine v2.4.0, Copyright (c) 1998-2013 Zend Technologies How can I select another php default/base version for the shell system

How do I change the default PHP base-version on shell?

允我心安 提交于 2021-02-02 09:07:04
问题 I have Centos 7 and Virtualmin installed, with the tipycal php-fpm 5.4, 7.0, 7.1 that you can choice between the versions you prefer on every virtualhost via Virtualmin control panel, and everything works well. But when I access to the server via SSH and check php -v I get this: PHP 5.4.16 (cli) (built: Oct 30 2018 19:30:51) Copyright (c) 1997-2013 The PHP Group Zend Engine v2.4.0, Copyright (c) 1998-2013 Zend Technologies How can I select another php default/base version for the shell system

Hitting hard memory limit when loading large image, possible to downsample while loading?

僤鯓⒐⒋嵵緔 提交于 2021-01-29 19:29:01
问题 I need to generate small thumbnails from potentially gigantic images (10,000 x 10,000). I'm using ImageMagick and PHP running in Google App Engine. My GAE has a hard memory limit of 512 MB. I've been trying to read the documentation for the imagemagick php extension, but the docs are pretty thin. I found this SO post that has an answer from the maintainer and tried to adapt their code. I tried to set the resource area so that images which were larger than 2000x2000 pixels would get swapped

PHP Sqlsrv connection not working when specify port, instance

半城伤御伤魂 提交于 2021-01-29 13:56:49
问题 PHP version:7.4, SQL Server version:2019, I was going to try to specify port in connection but it didn't work. <?php $serverName = "MYPCNAME, 1433"; $connectionInfo = array( "Database"=>"MyDB", "UID"=>"sa", "PWD"=>'MyPassword'); $conn = sqlsrv_connect( $serverName, $connectionInfo); if( $conn ) { echo "Connection established."; }else{ echo "Connection could not be established.<br />"; die( print_r( sqlsrv_errors(), true)); } ?> Above showed below message. [Microsoft][ODBC Driver 17 for SQL

Force 'fetch joined' relations to include IDENTITY of their ManyToOne relations using HYDRATE_ARRAY?

风流意气都作罢 提交于 2021-01-29 10:58:19
问题 I have a query in which I'm joining a number of tables to my original Person entity. A Person may have multiple Child relations (OneToMany), and a Child may have a School they go to (ManyToOne). Problem is, I don't need the entire School entity that connects to each child, only their id , which is already stored on Child . I'm using a Paginator to iterate through the results and I use HYDRATE_ARRAY to reduce overhead of the ORM parsing data to entity objects. But the id fields of unfetched

Laravel 5 relationship using a 'JSON' column

僤鯓⒐⒋嵵緔 提交于 2021-01-29 09:13:34
问题 Is it possible to use Eloquent to create a relationship in Laravel 5 where the foreign key exists in a field in a JSON column? If it is, how so? Example: I have a table called chats with a participantIds column, of a JSON datatype. The JSON format of the data looks like this: {"creator": "1", "recipient": "2"} I want to join the users table using those fields to get the participants of the Chats. How do I do that? 回答1: Laravel has no native support for JSON relationships. I created a package

How to generate unique tag_id start from F-00001 in php

你说的曾经没有我的故事 提交于 2021-01-29 05:00:55
问题 How to generate unique tag_id start from F-00001 when insert date into item table using php and mysql. This is my my table. I have three columns in my database table using MySQL. CREATE TABLE item ( id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, Name VARCHAR(30) NOT NULL, tag_id VARCHAR(8) NOT NULL ) 回答1: As @mickmackusa points out, you don't need to have this field in your database as you can automatically generate it from your auto-increment id value. There are a number of ways to do this.

Cannot update to php 5 to 7 in linux mint?

柔情痞子 提交于 2021-01-29 03:48:11
问题 I just want to update PHP 5 to PHP 7 and I got below error after adding this line " sudo add-apt-repository ppa:ondrej/php " Adding PPAs is not supported I am using Linux mint 17 and my web server is Nginx. How to fix this? Is there any alternative 回答1: To use the add-apt-repository you need to install software-properties-common or python-software-properties , just run the following command: sudo apt-get install software-properties-common python-software-properties Then run: sudo add-apt

Cannot update to php 5 to 7 in linux mint?

狂风中的少年 提交于 2021-01-29 03:44:09
问题 I just want to update PHP 5 to PHP 7 and I got below error after adding this line " sudo add-apt-repository ppa:ondrej/php " Adding PPAs is not supported I am using Linux mint 17 and my web server is Nginx. How to fix this? Is there any alternative 回答1: To use the add-apt-repository you need to install software-properties-common or python-software-properties , just run the following command: sudo apt-get install software-properties-common python-software-properties Then run: sudo add-apt

How to install imagick on PHP7 and Debian?

拟墨画扇 提交于 2021-01-28 21:52:55
问题 I need to install Imagick, but I've php7 and debian. I tried to do this : apt-get install php-imagick But I've this error : Unable to locate package php-imagick I also tried to do this apt-get install php5-imagick But O think this command is to php5, an dnow I've in /etc/php5/mods-available/ un imagick.ini So is imagick is install for php7? 回答1: It seems Imagick is not yet released for PHP7 on Debian stable. 回答2: Debian Buster and PHP7.3 sudo apt-get install -y libmagickwand-dev sudo apt-get