php 数组组合问题$array_old = array(
array(
'name' = 'param1',
'isNull' = 1,
'type' = 'int',
'value' = 1,
),
array(
'name' = 'param2',
'isNull' = 0,
'type' = 'string',
'value' = 'test',
),
);
php中的两组数如何相间组合到一起1.你先选择你想要组合的图形 2.右击选择弹出的“组合图形” 3.保存文件 。
php 如何计算数组中数字的组合?$arr = array(1,2,2,3,3,3,4,5,3,2,3,3);$num = count(array_count_values($arr));// 或者$num = count(array_unique($arr));
PHP问题: PHP二维数据组合1:
[id] = 408
[type] = china
[reality] = 120.00
[days] = 2016-06-20
2:
[id] = 408
[totalsave] = 200.00
[days] = 2016-06-20
3:
[id] = 408
[totalsave] = 3800.00
[days] = 2016-06-20
4:
[id] = 408
[totalwithdraw] = 92.00
[should] = 0.7176
[days] = 2016-06-20
【php数据组合 php数组相加组合数据】5:
[id] = 408
[totalwithdraw] = 3908.00
[should] = 14.8504
[days] = 2016-06-20
结论
[id] = 408
[totalwithdraw] = 82.00
[should] = 0.6396
[days] = 2016-06-20
[totalsave] = 3800.00
[type] = 中文
[reality] = 0.00
看不出有神马逻辑关系 。。
PHP 多个数组的排列组合$a = array(1,2,3,4,5,6,7);
$b = array(1,2,3);
$c = array(1,2,3,4,5,6,7,8,9,10);
$d = array($a,$b,$c);
//计算每一个数组的长度
$len = 1;
$arrLen = count($d); //需要排列数组有多少个
$recIndex = null; //记录当前该取的位置
//foreach 计数
$count_3 = 0;
foreach ($d as $key = $value) {
$lenRec[$count_3] = count($value);
$len = $lenRec[$count_3]*$len;
$recIndex[] = 0;//第一次全部取第0个
$count_3;
}
//算出% 的值
$count = 1;
foreach ($lenRec as $key = $value) {
$moduloVal = 1;
if($arrLen == $count){
$modulo[] = count($d[$arrLen-1]); //等于最后一个的长度
}else{
$count_1 = 1;
foreach ($lenRec as $index = $item) {
$count_1$count$moduloVal = $moduloVal*$item;
$count_1;
}
$modulo[] = $moduloVal;
}
$count;//为了防止$d key是有值的 不是自然序列 需要计数
}
$i = 1;
while ( $i = $len) {
$html = '';
$count_2 = 0;// 取模
$temp = '';
foreach ($d as $value) {
$html .= $value[$recIndex[$count_2]%$lenRec[$count_2]].",";
$count_2;
}
echo $html."br";
foreach ($modulo as $key = $value) {
if($i%$value =https://www.04ip.com/post/= 0$key$arrLen - 1 ){
$recIndex[$key] = $recIndex[$key]1;
}
if($key == $arrLen - 1){
if($i%$value =https://www.04ip.com/post/= 0){
$recIndex[$key] = 0;
}else{
$recIndex[$key] = $recIndex[$key]1;
}
}
}
$i;
//改变获取的位置
}
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