syntax-error

CakePHP application displays Syntax error, unexpected [

由 橙三吉。 提交于 2019-12-24 15:42:07
问题 This is the error I'm receiving: Error: syntax error, unexpected '[' Line: 10 I'm running my cakephp app on a linux server ubuntu 3.7, it's cakephp 2.3.7 and PHP 5.3.1. Now, I'm running WAMP on EC2 after installing linux. On my localmachine I run XAMPP on Windows 7, and it does not get the same error. This is the code where it displays error: 10: <?php foreach ($this->Session->read('Customer')['Addresses'] as $key => $value) { 11: $ids[$z++] = $value['id']; 12: ?> ... Since it does not give

I get a syntax error after JQuery upgrade from 1.51 to 2

由 风格不统一 提交于 2019-12-24 15:27:05
问题 The code hasn't changed only the JQuery version. A simplified version of the ajax call is: $.ajax( { type: 'GET', url: MapPath($(this).attr('path')), cache: false, data: '{}', dataType: 'html', success: function (result) { console.log(result); if ($(result).filter('#feedback').length > 0) { $('#center').children(':first').before($(result).filter('#feedback').outerHTML()); } else { $('#feedback').remove(); $('#dialog').html(result); $('#dialog').dialog('option', 'title', 'Edit Mailbox'); $('

Calling “del” on result of Function call results in “SyntaxError: can't delete function call”

由 ↘锁芯ラ 提交于 2019-12-24 13:28:42
问题 Consider below example: random_tile = random.choice(tiles) del random_tile It first assigns a random element from the list tiles to a variable, and then calls a function on that variable. Then if we want to shorten the code as follows: del random.choice(tiles) We would get a SyntaxError: can't delete function call . I tried eval() with no luck. How can this be solved? EDIT: What I am trying to do is to remove a element from the tiles list at random . I thought I found a more compact way of

mysqli_multi_query fails on multiple inserts

由 别说谁变了你拦得住时间么 提交于 2019-12-24 12:15:10
问题 Greetings, I'm having the following issue: When I attempt to execute several INSERT INTO queries using the mysqli_multi_query function, none of them are executed and I receive a standard "You have an error in your SQL syntax" error, but when I take the exact same string I passed to the function and paste into PHPMyAdmin and execute it, it works flawlessly! Here are the queries I am attempting to execute: INSERT INTO production VALUES( 120, 103, 10, 0, 0 ); INSERT INTO production VALUES( 120,

MySQL Zend Framework - SQLSTATE[42000]: Syntax error or access violation: 1064

由 本秂侑毒 提交于 2019-12-24 08:18:36
问题 I've read every response I could fine on SO before posting this question. Although similar, none addressed my particular problem (or I didn't recognize them doing so). I have a table class that extends Zend_Db_Table_Abstract. In the model, I'm trying to return a single row using a join() method and based on the table ID like this: $getCategoryResults = $this->select(); $getCategoryResults->setIntegrityCheck(false) ->from(array('c'=> 'categories', '*')) ->join(array('e' => 'events'),'c.events

MySQL Zend Framework - SQLSTATE[42000]: Syntax error or access violation: 1064

由 久未见 提交于 2019-12-24 08:17:14
问题 I've read every response I could fine on SO before posting this question. Although similar, none addressed my particular problem (or I didn't recognize them doing so). I have a table class that extends Zend_Db_Table_Abstract. In the model, I'm trying to return a single row using a join() method and based on the table ID like this: $getCategoryResults = $this->select(); $getCategoryResults->setIntegrityCheck(false) ->from(array('c'=> 'categories', '*')) ->join(array('e' => 'events'),'c.events

Alert function not working in coffeescript

由 穿精又带淫゛_ 提交于 2019-12-24 07:03:13
问题 I found about CoffeeScript in a blog and decided to give it a try ,my first project/code with it was this alert "Hello CoffeeScript!" It doesn't work and gives this reply ReferenceError: alert is not defined is there anything i am doing wrong? 回答1: JavaScript is a language which is strongly tied to the concept of environments . A browser and Node.js are two possible environments to run JS code (CoffeeScript compiles to JavaScript). When JavaScript is embedded in a browser, the global object

SyntaxError: Unexpected EOF

由 我的未来我决定 提交于 2019-12-24 06:53:29
问题 I add this code into my php footer: When I use the same syntax in an HTML document everything works well for me. <body> ... <script type="text/javascript"> // Browser detection function checkBrowserName(name){ var agent = navigator.userAgent.toLowerCase(); if (agent.indexOf(name.toLowerCase())>-1) { return true; } return false; } if(checkBrowserName('opera')){ /* load nothing */ } else if(checkBrowserName('')) { /* else if(checkBrowserName('firefox') || ('msie') || ('safari') || ('konqueror')

SyntaxError: Unexpected EOF

由 て烟熏妆下的殇ゞ 提交于 2019-12-24 06:53:08
问题 I add this code into my php footer: When I use the same syntax in an HTML document everything works well for me. <body> ... <script type="text/javascript"> // Browser detection function checkBrowserName(name){ var agent = navigator.userAgent.toLowerCase(); if (agent.indexOf(name.toLowerCase())>-1) { return true; } return false; } if(checkBrowserName('opera')){ /* load nothing */ } else if(checkBrowserName('')) { /* else if(checkBrowserName('firefox') || ('msie') || ('safari') || ('konqueror')

JavaScript Safari - SyntaxError: Unexpected token '>'

由 可紊 提交于 2019-12-24 06:48:25
问题 I have the problem that this JavaScript snippet runs in Firefox / Chrome without any problem, and Safari I get error: "SyntaxError: Unexpected token '>'". Here's the code: window.onclick = (test) => { const googleWindow = window.open(); fakeAjax(response => { googleWindow.location.replace(`https://google.com?q=${response}`); }); }; function fakeAjax(callback) { setTimeout(() => { callback('example'); }, 1); } I've googled and have already seen here in the forum, the problem there appears
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