Question

这是我写的代码，

import java.util.regex.*;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

String pStr ="Review Required";

int count = 50

for(int i=0;i<count;i++) 
{
String tmp = "wireReviewApprove:headerTable:0:wireStatusValue">\(\.\+\?\)<\/span>";
Pattern patternN = Pattern.compile(tmp);
Matcher matcher = patternN.matcher(pStr);

if (matcher.find())
    {
sampler.addArgument("selectedItemId["+i+"]","true");
    }
}

给出错误：ERROR - jmeter.util.BeanShellInterpreter：调用bsh方法时出错：eval源文件：内联评估：import java.util.regex.*; import java.util.regex.Matcher; import java.util.reg . . . '' Token Parsing Error: Lexical error at line 14, column 64. Encountered: "\\" (92), after : "" 2015/07/14 08:15:15 WARN - jmeter.modifiers.BeanShellPreProcessor: Problem in BeanShell script org.apache.jorphan.util.JMeterException: Error invoking bsh method: eval Sourced file: inline evaluation of: import java.util.regex。*; import java.util.regex.Matcher; import java.util.reg。。。＆＃39;＆＃39;令牌解析错误：第14行第64列的词汇错误。遇到：＆＃34; \＆＃34; （92），之后：＆＃34;＆＃34;

任何人都可以帮我解决这个问题吗？

Answer 1

int count = 50

转义tmp字符串的问题。如果原件看起来像：

wireReviewApprove:headerTable:0:wireStatusValue">\(\.\+\?\)<\/span>

你需要看起来像：

String tmp = "wireReviewApprove:headerTable:0:wireStatusValue\">\\(\\.\\+\\?\\)<\\/span>";

Full Beanshell代码：

int count = 50;

for (int i = 0; i < count; i++) {
    String tmp = "wireReviewApprove:headerTable:0:wireStatusValue\">\\(\\.\\+\\?\\)<\\/span>";        
    Pattern patternN = Pattern.compile(tmp);
    Matcher matcher = patternN.matcher(pStr);

    if (matcher.find()) {
        sampler.addArgument("selectedItemId[" + i + "]", "true");
    }
}

有关JMeter中Beanshell脚本的更多信息，请参阅How to use BeanShell: JMeter's favorite built-in component指南。

N.B。最好切换到JSR223 Sampler和Groovy语言，就像大负载一样，Beanshell可能成为瓶颈。

我想在Jemeter中使用beanshell预处理器执行正则表达式

1 个答案: