PHP简单的循环赋值问题
<code> echo $ss = count($row['key']); 输出是 111111 统计6个1 echo '<br>'; echo $num=count($ss); 统计输出只有1个1 </code>
如何实现 把统计输出6个1的值全部赋值到$num里全显示出来?
<code> for($i=1;$i</code>
这样列不出来,是不是什么地方写错了?
实际例子:
<code>$dosql->Execute("SELECT * FROM #@__infolist WHERE classid=2 AND delstate='' AND checkinfo=true ORDER BY orderid DESC LIMIT 0,100"); { while($row = $dosql->GetArray()) echo $ss = $row['keywords']; //输出 你好,你好,你好,不好,很好 echo $num= $ss; //输出 很好 //目标把 $ss 赋值给 $num 循环出来 for($i=1;$i</code>
回复内容:
<code> echo $ss = count($row['key']); 输出是 111111 统计6个1 echo '<br>'; echo $num=count($ss); 统计输出只有1个1 </code>
如何实现 把统计输出6个1的值全部赋值到$num里全显示出来?
<code> for($i=1;$i</code>
这样列不出来,是不是什么地方写错了?
实际例子:
<code>$dosql->Execute("SELECT * FROM #@__infolist WHERE classid=2 AND delstate='' AND checkinfo=true ORDER BY orderid DESC LIMIT 0,100"); { while($row = $dosql->GetArray()) echo $ss = $row['keywords']; //输出 你好,你好,你好,不好,很好 echo $num= $ss; //输出 很好 //目标把 $ss 赋值给 $num 循环出来 for($i=1;$i</code>
<code>$dosql->Execute("SELECT * FROM #@__infolist WHERE classid=2 AND delstate='' AND checkinfo=true ORDER BY orderid DESC LIMIT 0,100"); { $num = array(); while($row = $dosql->GetArray()) echo $ss = $row['keywords']; //输出 你好,你好,你好,不好,很好 echo $num= $ss; //输出 很好 //目标把 $ss 赋值给 $num 循环出来 $num[] = $row['keywords']; } var_dump($num);</code>
<code>$ss = count($row['key']); $num=count($ss) //这里$ss不是一个数组了,count(非数组)只会输出0(null时)和1</code>
第一次count完就是一个int型的数值。看不懂题主想要干嘛。
这里的 1 会默认做 int 处理的

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