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 <title><![CDATA[5个按钮，两个同时通报警怎么做？求最简单 ..]]></title>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261.html]]></link>
 <description><![CDATA[Latest 50 replies of 5个按钮，两个同时通报警怎么做？求最简单 ..]]></description>
 <copyright><![CDATA[Copyright(C) 工控人家园]]></copyright>
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 <title><![CDATA[]]></title>
 <description><![CDATA[每个按钮上升沿加1，下降沿减1，同一寄存器，大于等于2输出，应该可以实现吧]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1826268.html]]></link>
 <author><![CDATA[13670082332]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Sat, 15 Oct 2016 07:17:55 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[一楼的程序已经写的很清楚了]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1826253.html]]></link>
 <author><![CDATA[wycxks]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Sat, 15 Oct 2016 06:45:31 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[五个按钮　通为１　断为０　数据大于等于２就输出．]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1826186.html]]></link>
 <author><![CDATA[huiyang]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Sat, 15 Oct 2016 02:52:08 +0000]]></pubdate>
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<item>
 <title><![CDATA[]]></title>
 <description><![CDATA[LD&nbsp;&nbsp;M1
SUM&nbsp;&nbsp;K1X0 D0
LD&gt; D0 K1
OUT Y0
如果Y0是1时 M1断开变为0&nbsp;&nbsp; Y0保持是1还是变为0？]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1826166.html]]></link>
 <author><![CDATA[bank]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Sat, 15 Oct 2016 02:01:22 +0000]]></pubdate>
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<item>
 <title><![CDATA[]]></title>
 <description><![CDATA[1楼威武&#160;&#160;]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1825975.html]]></link>
 <author><![CDATA[apei2014]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 14 Oct 2016 05:56:21 +0000]]></pubdate>
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<item>
 <title><![CDATA[]]></title>
 <description><![CDATA[[s:21] 2楼正解]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1825915.html]]></link>
 <author><![CDATA[小小一电工]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 14 Oct 2016 03:00:18 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[大于等于吧]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1825910.html]]></link>
 <author><![CDATA[a470536252]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 14 Oct 2016 02:26:59 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[貌似没有说大于等于2个导通报警，只是说同时2个通报警。大于2个通不处理吗？]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1825899.html]]></link>
 <author><![CDATA[jeffwang2011]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Fri, 14 Oct 2016 01:36:49 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[[upload=1]]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1825807.html]]></link>
 <author><![CDATA[xingzb2]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Thu, 13 Oct 2016 11:39:16 +0000]]></pubdate>
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 <title><![CDATA[]]></title>
 <description><![CDATA[非常好]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1825803.html]]></link>
 <author><![CDATA[夏日雪梦]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Thu, 13 Oct 2016 11:14:03 +0000]]></pubdate>
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<item>
 <title><![CDATA[]]></title>
 <description><![CDATA[LD&#160;&#160; M8000
SUM&#160;&#160;K2X***&#160;&#160;D0
LD&gt;=&#160;&#160;D0 K2
OUT&#160;&#160; Y0&#160;&#160;
X***表示plc最大输入端子编号减掉4，]]></description>
 <link><![CDATA[http://www.ymmfa.com/read-gktid-1553261#1825802.html]]></link>
 <author><![CDATA[xingzb2]]></author>
 <category><![CDATA[综合讨论]]></category>
 <pubdate><![CDATA[Thu, 13 Oct 2016 11:06:19 +0000]]></pubdate>
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