请问一下 怎么取第一张图片的值呢？

Question

<?php$picarr = unserialize('a:8:{i:0;s:38:"uploads/image/20160706/1467745034.jpg,";i:1;s:38:"uploads/image/20160706/1467742197.jpg,";i:2;s:38:"uploads/image/20160706/1467738093.jpg,";i:3;s:38:"uploads/image/20160706/1467747796.jpg,";i:4;s:38:"uploads/image/20160706/1467742947.jpg,";i:5;s:38:"uploads/image/20160706/1467740461.jpg,";i:6;s:38:"uploads/image/20160706/1467746831.jpg,";i:7;s:38:"uploads/image/20160706/1467742639.jpg,";}');
foreach($picarr as $k)
{
$v = explode(',', $k);
//if($k==1) break;
echo $v[0];
}
?>
这里就全部都输出所有图片了。我只想要第一张图片的值。要怎么写。谢谢了。

Answer 1

echo $v[0];
break;

Answer 2

echo后直接break
或者$url = array();
foreach($picarr as $k)
{
    $v = explode(',', $k);
    $url[] = $v[0];
}
echo $url[0];

Answer 3

可以了。谢谢哥们 break

Answer 4

break  具体的用法是怎么用。能指导下嘛？

Answer 5

$picarr = unserialize('a:8:{i:0;s:38:"uploads/image/20160706/1467745034.jpg,";i:1;s:38:"uploads/image/20160706/1467742197.jpg,";i:2;s:38:"uploads/image/20160706/1467738093.jpg,";i:3;s:38:"uploads/image/20160706/1467747796.jpg,";i:4;s:38:"uploads/image/20160706/1467742947.jpg,";i:5;s:38:"uploads/image/20160706/1467740461.jpg,";i:6;s:38:"uploads/image/20160706/1467746831.jpg,";i:7;s:38:"uploads/image/20160706/1467742639.jpg,";}');
echo $picarr[0];uploads/image/20160706/1467745034.jpg,break 是跳出循环