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jQuery Ajax POST example with PHP
...me local variables
var $form = $(this);
// Let's select and cache all the fields
var $inputs = $form.find("input, select, button, textarea");
// Serialize the data in the form
var serializedData = $form.serialize();
// Let's disable the inputs for the duration of the Ajax ...
Set breakpoint in C or C++ code programmatically for gdb on Linux
How can I set a breakpoint in C or C++ code programatically that will work for gdb on Linux?
6 Answers
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Why can't (or doesn't) the compiler optimize a predictable addition loop into a multiplication?
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The compiler can't generally transform
for (int c = 0; c < arraySize; ++c)
if (data[c] >= 128)
for (int i = 0; i < 100000; ++i)
sum += data[c];
into
for (int c = 0; c < arraySize; ++c)
if (data[c] >= 12...
Retain cycle on `self` with blocks
...nswered Dec 4 '10 at 8:14
Lily BallardLily Ballard
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JavaScript open in a new window, not tab
I have a select box that calls window.open(url) when an item is selected. Firefox will open the page in a new tab by default. However, I would like the page to open in a new window, not a new tab.
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How do I write a short literal in C++?
...rt enough to compile this as if it's a short literal (i.e. it wouldn't actually allocate an int and then cast it every time).
The following illustrates how much you should worry about this:
a = 2L;
b = 2.0;
c = (short)2;
d = '\2';
Compile -> disassemble ->
movl $2, _a
movl $2, _b
m...
How do I find the duplicates in a list and create another list with them?
... @Hugo, to see the list of duplicates, we just need to create a new list called dup and add an else statement. For example: dup = [] else: dup.append(x)
– Chris Nielsen
Apr 29 '16 at 16:45
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Why does javascript replace only first instance when using replace? [duplicate]
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You need to set the g flag to replace globally:
date.replace(new RegExp("/", "g"), '')
// or
date.replace(/\//g, '')
Otherwise only the first occurrence will be replaced.
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php出现 Notice: Undefined index: xxx 的解决方法 - 更多技术 - 清泛网 - 专注IT技能提升
...方法供参考：
1、 修改 php.ini 文件中
error_reporting = E_ALL
为
error_reporting = E_ALL & ~E_NOTICE # 除去Notice警告
2、使用代码
ini_set( 'error_reporting', E_ALL ^ E_NOTICE );
ini_set( 'display_errors', '0' );
3、使用@抑制错误
@$name = $_GET['name'];
php出现 Notice: Undefined index: xxx 的解决方法 - 更多技术 - 清泛网 - 专注IT技能提升
...方法供参考：
1、 修改 php.ini 文件中
error_reporting = E_ALL
为
error_reporting = E_ALL & ~E_NOTICE # 除去Notice警告
2、使用代码
ini_set( 'error_reporting', E_ALL ^ E_NOTICE );
ini_set( 'display_errors', '0' );
3、使用@抑制错误
@$name = $_GET['name'];