不得不将PHP代码部署到只有PHP5.2.8的共享服务器上.对于preg_filter()在5.3+中添加的调用,所有代码的工作都是EXCEPT (代码编写时也是如此).
任何人都可以preg_filter()在PHP5.2中推荐一个简单的替代品吗?
我是php的新手.我需要一些帮助.我有一个阵列
Array ( [_] => Array ( [0] => [1] => )
[123_] => Array ( [0] => 123 [1] => )
[1234_] => Array ( [0] => 1234 [1] => )
)
Array ( [_] => Array ( [0] => [1] => )
[12345_] => Array ( [0] => 12345 [1] => )
[1234_] => Array ( [0] => 1234 [1] => )
)
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所以..我的问题是我想要一个包含所有这些键和值的数组
Array ( [_] => Array ( [0] => [1] => )
[123_] => Array …Run Code Online (Sandbox Code Playgroud) 查询:
INSERT INTO Customer2
VALUES (1, 'Mrs','Jill','Hill','2 Step St','Hillington','Uxbridge',
'Middx''UB10 8XY','020 8999 684') ;
Run Code Online (Sandbox Code Playgroud) 我想将块大小转换为MB.我/e在替换中使用了该选项.当我在替换部分添加起始MB时,它给我错误.
例如:
这有效.
echo "16777216 SELECT" |perl -lane 's#(\d+)(\s+SELECT)#$1/(1024*1024*2)#e; print'
8
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这给了我错误.
echo "16777216 SELECT" |perl -lane 's#(\d+)(\s+SELECT)#$1/(1024*1024*2) MB $2#e; print'
Bareword found where operator expected at -e line 1, near ") MB"
(Missing operator before MB?)
syntax error at -e line 1, near ") MB "
Execution of -e aborted due to compilation errors.
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任何修复第二个的帮助?
这是我关于 setter 和 getter 的任务,由于某种原因它不起作用。谁能检查一下我的问题是什么?谢谢你。
public class FlightTest
{
public static void main (String [] args)
{
String name;
String number;
String Orig;
String Desti;
Scanner scan = new Scanner (System.in);
Flight data = new Flight ();
System.out.print ("Airline Name: ");
String AirlineName = scan.nextLine ();
data.setAirlineName (name);
System.out.print ("Flight Number: ");
String FlightNumber = scan.nextLine ();
data.setFlightNumber (number);
System.out.print ("Origin: ");
String Origin = scan.nextLine();
data.setOrigin (Orig);
System.out.print ("Destination: ");
String Destination = scan.nextLine ();
data.setDestination (Desti);
System.out.println (data); …Run Code Online (Sandbox Code Playgroud) 我想从以下字符串中捕获:
var MyCode = "jdgsrtjd";
var ProductId = 'PX49EZ482H';
var TempPath = 'Media/Pos/';
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我想得到的是单引号ProductId值之间的可变长度值
PX49EX482H
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我有这个,我认为它很接近,但是单引号让我感到沮丧.我不确定如何正确逃脱它们.
preg_match('/var ProductID ='(.*?)';/', $str, $matches);
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提前致谢!
编译此代码时出现以下消息.
Question2中的ExtractChars(java.lang.String,int)无法应用于()
我该怎么办?
谢谢.
import java.util.Scanner;
public class Question2
{
public static void main (String[] args)
{
ExtractChars();
}
public static String ExtractChars(String sOrg, int startPos)
{
Scanner scan = new Scanner (System.in);
String value = "";
System.out.print("Enter a string: ");
sOrg = scan.next();
System.out.print("/nEnter an integer: ");
startPos = scan.nextInt();
if (sOrg.length()<=startPos)
{
value = "";
}
else
{
for (int i=startPos; i<sOrg.length();i=i+2)
{
value = value + sOrg.charAt(i);
}
}
return value;
}
}
Run Code Online (Sandbox Code Playgroud) 我目前正在使用这个正则表达式^[A-Z0-9 _]*$来接受字母,数字,空格和下划线.我需要修改它以允许.
大家好我已经看了很多PHP脚本,我看到了这样的东西
$variable->function($var1, $var2, $var3)
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那究竟->是做什么的呢?
谢谢你
我的LCM计划结果出错了.
如果找到数字的gcd,然后用gcd划分产品.
int gcd(int x, int y)
{
while(y != 0)
{
int save = y;
y = x % y;
x = save;
}
return y;
}
int lcm(int x, int y)
{
int prod = x * y;
int Gcd = gcd(x,y);
int lcm = prod / Gcd;
return lcm;
}
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任何帮助非常感谢.