我有这两个数组:
$ar1 = array(
0 => array(
'name' => 'MIRIAN',
'total' => '34'
),
1 => array(
'name' => 'THIAGO',
'total' => '29'
),
2 => array(
'name' => 'EDUARDO',
'total' => '3'
)
);
$ar2 = array(
0 => array(
'operator' => 'THIAGO',
'totalop' => '703'
),
1 => array(
'operator' => 'MIRIAN',
'totalop' => '180'
)
);
我想创建第三个像这样的人,但我不知道该怎么办:
$ar3 = array(
0 => array(
'name' => 'MIRIAN',
'total' => '34'
'totalop' => '180'
),
1 => array(
'name' => 'THIAGO',
'total' => '29',
'totalop' => '703'
),
2 => array(
'name' => 'EDUARDO',
'total' => '3'
)
);
这样做的最佳方法是什么,我尝试了很多方法,但没有成功?
感谢您的帮助
答案 0 :(得分:2)
我将通过示例完成以前的答案:)
<?php
$ar1 = array(
'#MIRIAN' => array(
'name' => 'MIRIAN',
'total' => '34'
),
'#THIAGO' => array(
'name' => 'THIAGO',
'total' => '29'
),
'#EDUARDO' => array(
'name' => 'EDUARDO',
'total' => '3'
)
);
$ar2 = array(
'#THIAGO' => array(
'operator' => 'THIAGO',
'totalop' => '703'
),
'#MIRIAN' => array(
'operator' => 'MIRIAN',
'totalop' => '180'
)
);
?>
我修改了$ar1
和$ar2
:我添加了一个id键(使用任何唯一值)。有时这只是编辑id并避免使用大算法这么简单:)
<?php var_dump( array_merge($ar1, $ar2) ); ?>
结果 - 你放松了第二个暗淡:
array (size=3)
'#MIRIAN' =>
array (size=2)
'operator' => string 'MIRIAN' (length=6)
'totalop' => string '180' (length=3)
'#THIAGO' =>
array (size=2)
'operator' => string 'THIAGO' (length=6)
'totalop' => string '703' (length=3)
'#EDUARDO' =>
array (size=2)
'name' => string 'EDUARDO' (length=7)
'total' => string '3' (length=1)
<?php var_dump( array_merge_recursive($ar1, $ar2) ); ?>
预期结果! (但我们有两个名字/经营者)
array (size=3)
'#MIRIAN' =>
array (size=4)
'name' => string 'MIRIAN' (length=6)
'total' => string '34' (length=2)
'operator' => string 'MIRIAN' (length=6)
'totalop' => string '180' (length=3)
'#THIAGO' =>
array (size=4)
'name' => string 'THIAGO' (length=6)
'total' => string '29' (length=2)
'operator' => string 'THIAGO' (length=6)
'totalop' => string '703' (length=3)
'#EDUARDO' =>
array (size=2)
'name' => string 'EDUARDO' (length=7)
'total' => string '3' (length=1)
<?php
$return = $ar1;
foreach( $ar2 as $k2 => $v2 ) {
$isFoundKey = null;
foreach( $return as $k2_2 => $v2_2 ) {
if( isset($v2['operator'], $v2_2['name']) && $v2['operator'] == $v2_2['name'] )
$isFoundKey = $k2_2;
}
if( !is_null($isFoundKey) )
$return[$isFoundKey] = array_merge($return[$isFoundKey], $v2);
else
$return[$k2] = $v2;
}
var_dump($return);
?>
结果:
array (size=3)
'#MIRIAN' =>
array (size=4)
'name' => string 'MIRIAN' (length=6)
'total' => string '34' (length=2)
'operator' => string 'MIRIAN' (length=6)
'totalop' => string '180' (length=3)
'#THIAGO' =>
array (size=4)
'name' => string 'THIAGO' (length=6)
'total' => string '29' (length=2)
'operator' => string 'THIAGO' (length=6)
'totalop' => string '703' (length=3)
'#EDUARDO' =>
array (size=2)
'name' => string 'EDUARDO' (length=7)
'total' => string '3' (length=1)
现在,您只需要清理数据:)(仅保留名称或运算符 - 请参阅@ofca答案)
答案 1 :(得分:0)
通过使用$ar3
按名称重新编号$ar1
来创建array_column
。然后迭代$ar2
并将totalop
值附加到$ar3
中的相应键。
$ar3 = array_column($ar1, null, 'name');
foreach ($ar2 as $x) {
$ar3[$x['operator']]['totalop'] = $x['totalop'];
}
答案 2 :(得分:-1)
解决方案:
$ar1 = array(
0 => array(
'name' => 'MIRIAN',
'total' => '34'
),
1 => array(
'name' => 'THIAGO',
'total' => '29'
),
2 => array(
'name' => 'EDUARDO',
'total' => '3'
)
);
$ar2 = array(
0 => array(
'operator' => 'THIAGO',
'totalop' => '703'
),
1 => array(
'operator' => 'MIRIAN',
'totalop' => '180'
)
);
$index = [];
$ar1Key = 'name';
$ar2Key = 'operator';
foreach ($ar1 as $item) {
$index[$item[$ar1Key]] = $item;
}
foreach ($ar2 as $item) {
$pk = $item[$ar2Key];
unset($item[$ar2Key]);
if (isset($index[$pk])) {
$index[$pk] = array_merge($index[$pk], $item);
} else {
$index[] = $item;
}
}
$ar3 = array_values($index);
print_r($ar3);