如何在循环外循环时移动

I have a much more complex problem that I am trying to resolve, however it all comes down to getting mysql_fetch_array values out of a while loop so I can use the values for comparison somewhere else.

我有一个更复杂的问题，我试图解决，但这一切都归结为从while循环中获取mysql_fetch_array值，所以我可以在其他地方使用这些值进行比较。

For example:

例如：

 $result = mysql_query("SELECT name FROM Names");
 while ($data = mysql_fetch_array($result))
 {
   $val = $data['name'];
   //let's say there are 5 names here//
   $array1 = array($val);
 }

 //now, I would like to use the five names outside of the loop//

I do not understand how to get all five names out of this loop. I have them in an array, but when I print the array from outside the loop, all I get is the last record. If I print the array from within the Loop, I get all records.

我不明白如何从这个循环中取出所有五个名字。我把它们放在一个数组中，但是当我从循环外部打印数组时，我得到的只是最后一条记录。如果我从Loop中打印数组，我会得到所有记录。

Is it possible to store the values in the loop, move them from the loop, and then use them somewhere else?

是否可以将值存储在循环中，将它们从循环中移出，然后在其他地方使用它们？

4 个解决方案

#1

You could do somthing like this:

你可以这样做：

 $result = mysql_query("SELECT name FROM Names");
 $dataset = array();
 while ($data = mysql_fetch_array($result))
 {
   $dataset[] = $data;
 }

Then you are free to use $dataset outside the loop...

然后你可以在循环外使用$ dataset ...

#2

$array1 = array();
$result = mysql_query("SELECT name FROM Names");
while ($data = mysql_fetch_array($result)){
 $val = $data['name'];
  //let's say there are 5 names here//
  $array1[] = $val;
}

#3

You are setting $array1 every time in the loop. You need to ADD $val to the array, not set the array to $val.

你每次都在循环中设置$ array1。您需要向数组添加$ val，而不是将数组设置为$ val。

$result = mysql_query("SELECT name FROM Names");
while ($data = mysql_fetch_array($result))
{
    $val = $data['name'];
    $array1[] = array($val);
}

Notice the $array1[] = instead of $array1 =.

注意$ array1 [] =而不是$ array1 =。

#4

You are overwriting the array. Everytime the loop execute it overwrites the value, so you get only the last value. Try

你正在覆盖数组。每次循环执行时都会覆盖该值，因此您只获得最后一个值。尝试

$result = mysql_query("SELECT name FROM Names");
$i = 0;
while ($data = mysql_fetch_array($result))
$val = $data['name'];
$array1[$i] = $val;
$i++;
}

#1