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I am trying to create fake IP address using the following javascript random method. But not works.

any one help me to get the correct way to do this?

here is my try: I am looking like ( "192.168.10.2")

 console.log( Math.php yui手工制作rss阅读器– 某人的栖息地 Ajax php yui手工制作rss阅读器– Ajax random(256) + "." + Math.random(256)Fakeid idsfun Twitter Idsfun Online Fakeid Idsfun EqR1a0 + "."php Ajax Ajax yui手工制作rss阅读器– php 某人的栖息地 yui手工制作rss阅读器– +Ajax yui手工制作rss阅读器– php php 某人的栖息地 yui手工制作rss阅读器– Ajax Mathyui手工制作rss阅读器– php php Ajax yui手工制作rss阅读器– Ajax 某人的栖息地 .random(256) + "."Ajax yui手工制作rss阅读器– php php 某人的栖息地 yui手工制作rss阅读器– Ajax + Math.random(256) );
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marked as duplicate by John3136, JJJ, T.J. Crowder Apr 18 '17 at 5:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

7

The code should be :

var ip = 某人的栖息地 yui手工制作rss阅读器– yui手工制作rss阅读器– Ajax Ajax php php (Math.floor(Math.random() * 255) + 1)+"."+(Math.floor(Math.random() *yui手工制作rss阅读器– 某人的栖息地 Ajax php php yui手工制作rss阅读器– Ajax 255) + 0)+"."+(Math.floor(Math.random() *yui手工制作rss阅读器– Ajax yui手工制作rss阅读器– php php 某人的栖息地 Ajax php Ajax yui手工制作rss阅读器– php Ajax 某人的栖息地 yui手工制作rss阅读器– 255) + 0)+"."+(Math.floor(Math.random() * 255yui手工制作rss阅读器– Ajax 某人的栖息地 php Ajax php yui手工制作rss阅读器– ) + yui手工制作rss阅读器– 某人的栖息地 php php Ajax yui手工制作rss阅读器– Ajax 0); console.log(ip)
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The code (Math.floor(Math.random() * 255) + 1) means generate a random number between 1 and 255.

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