How to use PowerShell select-string to find string and then copy right part of that string until anoter string begins

Question

I need to make a script that will find the "Error" in the log file and after this error copy its description, which is in a separate string. I manage to find and copy the entire error string, but it turns out that the block with the error description is located a little lower and I can not copy it.

Select-String C:\TEMP\Skript\Log.xml -Pattern ' level="ERROR" '  -AllMatches  |  select | out-file C:\TEMP\Skript\LogErrors.xml -append

How do I specify additional parameters for a search?

The comment is in a separate block and not on the same line with the error. This block is not necessarily 1 line long.

<log4j:event logger="COMP" timestamp="1579174503853" level="ERROR" thread="1"><log4j:message>
here comes
a long error discription 
for a few lines.
</log4j:message>

This is where the error description ends. So i must find ERROR in log and then copy this string and the description.

Expected Result: Get an every error with its description, even if the error description takes several lines.

   <log4j:event logger="COMP" timestamp="1579174503853" level="ERROR" thread="1"><log4j:message>
here comes a long error discription  for a few lines.
 </log4j:message>

"located a little lower", you mean on the next line? Use -Context 0,1 with Select-String to grab one line after the match — Mathias R. Jessen
– Mathias R. Jessen, Commented Jan 17, 2020 at 11:10
I mean, the comment is in a separate block and not on the same line with the error. This block is not necessarily 1 line long. <log4j:event logger="COMP" timestamp="1579174503853" level="ERROR" thread="1"><log4j:message> here comes a long error discription for a few lines. </log4j:message> This is where the error description ends. So i must find ERROR in log and then copy this string and the description. — Eugen Calavera
– Eugen Calavera, Commented Jan 17, 2020 at 11:49
You could use the same approach I suggested in another question here. — Lieven Keersmaekers
– Lieven Keersmaekers, Commented Jan 17, 2020 at 11:59
Please update your question (don't use comments) to show sample input and expected output. — mklement0
– mklement0, Commented Jan 18, 2020 at 15:32

RagdGC · Accepted Answer · 2020-01-24 02:41:24Z

Considering the information in file Log.xml is:

    <log4j:event logger="COMP" timestamp="1579174503853" level="INFO" thread="1"><log4j:message>
    Information message
    </log4j:message>
    <log4j:event logger="COMP" timestamp="1579174503853" level="ERROR" thread="1"><log4j:message>
    here comes
    a long error discription 
    for three few lines.
    </log4j:message>
    <log4j:event logger="COMP" timestamp="1579174503853" level="INFO" thread="1"><log4j:message>
    Information message
    </log4j:message>
    <log4j:event logger="COMP" timestamp="1579174503855" level="ERROR" thread="1"><log4j:message>
    here comes a second long error discription 
    for a few lines.
    </log4j:message>
    <log4j:event logger="COMP" timestamp="1679174503855" level="ERROR" thread="1"><log4j:message>
    single line error. 
    </log4j:message>
    <log4j:event logger="COMP" timestamp="1579174503853" level="INFO" thread="1"><log4j:message>
    Information message
    </log4j:message>

You can use the next command

    Get-Content .\Log.xml | foreach {
    if ($_.Contains('level="ERROR"')) {
    $errorFound= $true }
    if ($errorFound -And $_.StartsWith('<l')) { 
    $linea =""; $_ } 
    elseif($errorFound -And $_.StartsWith('</')) { 
    $linea + "`n" + $_ ; $errorFound = $false; } 
    elseif ($errorFound) { 
    $linea = $linea +  $_  }
    } | Add-Content LogError.xml

This will save the information with the format you want in the file LogError.xml

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