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How can I find where I will be redirected using cURL?
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This makes php follow the redirect. I dont want to follow the redirect, I just want to know the url of the redirected page.
– Thomas Van Nuffel
Aug 19 '10 at 8:50
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What are the recommendations for html tag?
...t;/a>
where ${uri} basically translates to $_SERVER['REQUEST_URI'] in PHP, ${pageContext.request.requestURI} in JSP, and #{request.requestURI} in JSF. Noted should be that MVC frameworks like JSF have tags reducing all this boilerplate and removing the need for <base>. See also a.o. What ...
AngularJS $http and $resource
...hBen Lesh
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1
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Listening for variable changes in JavaScript
...c.innerHTML = c.innerHTML + '<br />' + t;
}
// Demo
var myVar = 123;
Object.defineProperty(this, 'varWatch', {
get: function () { return myVar; },
set: function (v) {
myVar = v;
print('Value changed! New value: ' + v);
}
});
print(varWatch);
varWatch = 456;
pri...
通过 ulimit 改善系统性能 - 操作系统(内核) - 清泛网 - 专注C/C++及内核技术
...资源的条件下保证程序的运作，ulimit 是我们在处理这些问题时，经常使用的一种简单手段。ulimit 是一种 linux 系统的内键功能，它具有一套参数集，用于为由它生成的 shell 进程及其子进程的资源使用设置限制。本文将在后面的...
Implode an array with JavaScript?
Can I implode an array in jQuery like in PHP?
7 Answers
7
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Run cURL commands from Windows console
...WARNING: This does NOT pass GET parameters to the page. I used this with a PHP page. curl https://www.example.com/mypage.php?action=hello. In the mypage.php script, $_GET['action'] is empty
– Stephen R
Jun 26 '19 at 0:31
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Using curl to upload POST data with files
... file" \
-F "image=@/home/user1/Desktop/test.jpg" \
localhost/uploader.php
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How do I hide javascript code in a webpage?
...s say you are worried about exposing a secret. Let's say you put it into a PHP file and call it via Ajax. Then anyone can call that PHP file and find the secret. There is probably a way to protect secrets using PHP, and I've been struggling to find it. Generate a random number and require that all a...
jQuery loop over JSON result from AJAX Success?
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You can also use the getJSON function:
$.getJSON('/your/script.php', function(data) {
$.each(data, function(index) {
alert(data[index].TEST1);
alert(data[index].TEST2);
});
});
This is really just a rewording of ifesdjeen's answer, but I thou...