$string = '20110306';
$pattern = '(\d{6})(\d{2})';
$replacement = '$101';
echo preg_replace($pattern, $replacement, $string);
我希望它回应 20110301
我使用http://gskinner.com/RegExr/来提出搜索和替换模式,也许我在更换找到的模式时遗漏了一些东西.
它给了我以下警告:
消息:preg_replace()[function.preg-replace]:未知修饰符'('
我有一个名为"file1"的文本文件,其中包含以下数据:
apple
appLe
app^e
app\^e
现在给出的命令是:
1.)grep app[\^lL]e file1
2.)grep "app[\^lL]e" file1
3.)grep "app[l\^L]e" file1
4.)grep app[l\^L]e file1
output in 1st case : app^e
output in 2nd case :
apple
appLe
app^e
output in 3rd case :
apple
appLe
app^e
output in 4th case :
apple
appLe
app^e
为什么这样..？
请帮忙..!
我需要做以下选择:
select distinct type as option from pages
但我不断收到错误.你能帮我吗？
期望以下例程在单个链表的开头删除节点,但有时会失败.
void remove_from_front(node *start)
{
delete start;
start = start->link;
print_list(start);
}
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/ HTTP/1.1" 200 169 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/.treeinfo HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Fedora HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Server HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/Client HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/RedHat HTTP/1.1" 404 182 "-" "Python-urllib/2.6"
127.0.0.1 - - [08/Mar/2011:00:26:27 +0530] "HEAD /sk/CentOS HTTP/1.1" …我有以下问题
给定一个字符串,返回一个"清理"字符串,其中相同的字符相同的字符已减少为一个字符.所以"yyzzza"产量"yza".
stringClean("yyzzza") ? "yza"
stringClean("abbbcdd") ? "abcd"
stringClean("Hello") ? "Helo"
我正在尝试输入代码 stringClean("abbbcdd") ? "abcd"
我的代码在下面.我在进行相邻字符比较后得到部分附加字符串,因此截至现在我得到附加的stringBuilder "sb=abc",这不是正确的输出我应该得到输出"abcd",
class cleanString{
public static String stringClean(String str){
int startIndex = str.indexOf(str);
char startChar = '\u0000';
char adjacentChar = '\u0000';
System.out.println("startIndex-->" + startIndex);
final StringBuilder sb = new StringBuilder();
for(startIndex = 0; startIndex < str.length(); startIndex += 1){
startChar = str.charAt(startIndex);
System.out.println("startIndex ::" + startIndex);
System.out.println("startChar ::" + startChar);
final int adjacentPosition = startIndex + 1;
System.out.println("adjacentPosition …如果我有一个Java类,如下所示,相同的Perl表示是什么？
class Temp {
int my_int;
int my_array[];
Temp() {
}
};
具体来说,我有兴趣知道如何使用数组作为对象变量？
我是指针和链表的新手,但我正在尝试编写一个简单的程序,将文本文件中的数据读入链表.我输入功能有问题.它看起来像这样:
DVDNode* CreateList(string fileName)
{
ifstream inFile;
inFile.open(fileName.c_str());
DVDNode* head = NULL;
DVDNode* dvdPtr;
dvdPtr = new DVDNode;
while(inFile && dvdPtr != NULL)
{
getline(inFile, dvdPtr -> title);
getline(inFile, dvdPtr -> leadActor);
getline(inFile, dvdPtr -> supportingActor);
getline(inFile, dvdPtr -> genre);
cin >> dvdPtr -> year;
cin >> dvdPtr -> rating;
getline(inFile, dvdPtr -> synopsis);
dvdPtr -> next = head;
head = dvdPtr;
dvdPtr = new DVDNode;
}
delete dvdPtr;
inFile.close();
return head;
}
我也有一个输出函数,如下所示:
void OutputList(DVDNode* head, string outputFile)
{ …Array
(
[0] => '1 Fail'
[1] => '2 Fail'
[2] => '3 Pass'
[3] => '4 Pass'
[4] => '5 Pass'
)
Array
(
['1 Fail'] => '1 Fail'
['2 Fail'] => '2 Fail'
['3 Pass'] => '3 Pass'
['4 Pass'] => '4 Pass'
['5 Pass'] => '5 Pass'
)
是否有一个PHP函数从数组1转换为数组2
PS:我知道这一点,所以我正在寻找内置功能
foreach($result as $value)
{
$assoc[$value] = $value;
}