为啥这样的二维数组foreach后只有最后一个值输入

WBOY
Release: 2016-06-13 11:50:21
Original
933 people have browsed it

为什么这样的二维数组foreach后只有最后一个值输入?
$res = array(
array(
'goods_id' => 14,
'goods_number' => 1
),
array(
'goods_id' => 16,
'goods_number' => 1
),
array(
'goods_id' => 18,
'goods_number' => 1
)
);
上面数组,我想遍历出goods_id 从14 到 18,也就是14,15,16,17,18 ,从而得到goods_id 对应的 goods_number,如果goods_id不存在数组里,goods_number就给予0给它。

foreach($res as $value)
{
if($value['goods_id'] == 14)
{
echo $value['goods_number'];
echo "
";
}
else
{
echo 0;
echo "
";
}

if($value['goods_id'] == 15)
{
echo $value['goods_number'];
}
else
{
echo 0;
echo "
";
}

if($value['goods_id'] == 16)
{
echo $value['goods_number'];
}
else
{
echo 0;
echo "
";
}

if($value['goods_id'] == 17)
{
echo $value['goods_number'];
}
else
{
echo 0;
echo "
";
}

if($value['goods_id'] == 18)
{
echo $value['goods_number'];
}
else
{
echo 0;
echo "
";
}
}
为什么我这样foreach 不行呢?

我要的结果是:
goods_id:14 时 goods_number:1
goods_id:15 时 goods_number:0
goods_id:16 时 goods_number:1
goods_id:17 时 goods_number:0
goods_id:18 时 goods_number:1
所以应该echo的结果是
1
0
1
0
1

为什么我的foreach 得到的不是这样的结果呢?
------解决方案--------------------

$res = array(<br />	array(<br />		'goods_id' => 14,<br />		'goods_number' => 1<br />		),<br />	array(<br />		'goods_id' => 16,<br />		'goods_number' => 1<br />		),<br />	array(<br />		'goods_id' => 18,<br />		'goods_number' => 1<br />		)<br />	);<br />$newArr = array();<br />foreach ($res as $key => $value) {<br />	$newArr[$value['goods_id']] = $value['goods_number'];<br />}<br /><br />//var_dump(array_values($res));exit;<br />$range = range(14, 18);<br /><br />foreach ($range as $keys => $values) {<br /><br />	if(array_key_exists($values, $newArr)){<br />		echo 'goods_id:'.$values.' 时 goods_number:'.$newArr[$values];<br />		echo '<br />';<br />	}else{<br />	<br />		echo 'goods_id:'.$values.' 时 goods_number:0';<br />		echo '<br />';<br />	}<br />}
Copy after login

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template