我有两张桌子,“标题”和“马赫”:
头:
STATO |ID |CAMPIONATO
Italy |5 |Serie A
Spain |1 |Primera Division
France|2 |Coppe de France
马赫:
STATO |home |away |scoreH|scoreA|
Italy |juve |Milan |0 |0 |
Italy |Lazio |Roma |0 |1 |
Spain |Deportivo|Bilbao |1 |0 |
Spain |A Madrid |Sevilla |1 |2 |
France|Lille |Perigord |0 |0 |
我使用以下PHP代码来提取数据。我应该创建一个包含标题的数组,并且组匹配
$num=0;
$mach=-1;
foreach($first as $row2){ // cycling once
foreach($header as $row){ // cycles x times
.... //Here recover the header data
$stato;
$id;
$campionato
for($i=0;$i<$count;++$i){ // in $count know how many times I have to do the cycle for each header (stato)
$mach=$mach+1;
.... // Here recover the mach data
$home;
$away;
$scoreH;
$scoreA;
$info[$mach]= array('home'=>$home,'away'=>$away,'scoreH'=>$scoreH,'scoreA'=>$scoreA);
$groups[$num]=array($info[$mach]);
}//end for cycles
$headerArray[$num]=array('header'=>['stato'=>$stato,'id'=>$id,'campionato'=>$campionato],'mach'=>$groups[$um]);
$num++;
} //end header cycles
$blockArray[]=array('incontri'=>$headerArray);
} //end first cycle
print_r($blockArray);
问题在于返回所有日期匹配的$ groups。(意大利,西班牙,法国)不仅仅是标题组。 如果你使用这行PHP代码:
array_push ($groups,$info[$mach]);
而不是:
$Groups[$num] = array($info[$mach]);
只返回每个标题的最后一个匹配(循环)。
我想创建这个数组
['incontri'=>
[
header=>
['stato'=>Italy,'id'=>5,'campionato'=>Serie A],
mach=>
['home'=>Juve,'away'=>Milan,'scoreH'=>0,'scoreA'=>0],
['home'=>Lazio,'away'=>Roma,'scoreH'=>0,'scoreA'=>0]
],
[....],
]
我做错了什么?
答案 0 :(得分:0)
我无法对你的代码产生任何精神上的牵引力,所以我自己重写了一遍。 如果我理解你的目标,我已经精心设计了一个更精简,更清晰的过程来生成你想要的多维数组。
*有些注意事项:
$adeguata_mach_keys
搜索$mach
中查找与STATO
值匹配的header STATO
值的所有行,并返回键。以下是我测试成功的代码:
$first=array(); // only cycles once, so don't bother looping it, just call it by key (I am assuming 0, but that may be wrong)
$first[0]=array(
array("STATO"=>"Italy","ID"=>5,"CAMPIONATO"=>"Serie A"),
array("STATO"=>"Spain","ID"=>1,"CAMPIONATO"=>"Primera Division"),
array("STATO"=>"France","ID"=>1,"CAMPIONATO"=>"Coppe de France")
);
$mach=array(
array("STATO"=>"Italy","home"=>"juve","away"=>"Milan","scoreH"=>"0","scoreA"=>"0"),
array("STATO"=>"Italy","home"=>"Lazio","away"=>"Roma","scoreH"=>"0","scoreA"=>"1"),
array("STATO"=>"Spain","home"=>"Deportivo","away"=>"Bilbao","scoreH"=>"1","scoreA"=>"0"),
array("STATO"=>"Spain","home"=>"A Madrid","away"=>"Sevilla","scoreH"=>"1","scoreA"=>"2"),
array("STATO"=>"Spain","home"=>"Lille","away"=>"Perigord","scoreH"=>"0","scoreA"=>"0")
);
foreach($first[0] as $header_index=>$header_row){
foreach($header_row as $header_row_key=>$header_row_val){
$result[$header_index]["header"][$header_row_key]=$header_row_val;
// identify which mach rows hold matching STATO values
$adeguata_mach_keys=array_keys(array_intersect(array_column($mach,"STATO"),array($header_row["STATO"])));
foreach($adeguata_mach_keys as $mach_index=>$mach_key){
foreach($mach[$mach_key] as $mach_row_key=>$mach_row_val){
$result[$header_index]["mach"][$mach_index][$mach_row_key]=$mach_row_val;
}
}
}
}
$result=array("incontri"=>$result); // I wouldn't do this, but if it is necessary for your case, fine.
print_r($result);
这是结果(没有意外的子阵列覆盖):
array (
'incontri' => array (
0 => array (
'header' => array (
'STATO' => 'Italy',
'ID' => 5,
'CAMPIONATO' => 'Serie A',
),
'mach' => array (
0 => array (
'STATO' => 'Italy',
'home' => 'juve',
'away' => 'Milan',
'scoreH' => '0',
'scoreA' => '0',
),
1 => array (
'STATO' => 'Italy',
'home' => 'Lazio',
'away' => 'Roma',
'scoreH' => '0',
'scoreA' => '1',
),
),
),
1 => array (
'header' => array (
'STATO' => 'Spain',
'ID' => 1,
'CAMPIONATO' => 'Primera Division',
),
'mach' => array (
0 => array (
'STATO' => 'Spain',
'home' => 'Deportivo',
'away' => 'Bilbao',
'scoreH' => '1',
'scoreA' => '0',
),
1 => array (
'STATO' => 'Spain',
'home' => 'A Madrid',
'away' => 'Sevilla',
'scoreH' => '1',
'scoreA' => '2',
),
),
),
2 => array (
'header' => array (
'STATO' => 'France',
'ID' => 1,
'CAMPIONATO' => 'Coppe de France',
),
'mach' => array (
0 => array (
'STATO' => 'France',
'home' => 'Lille',
'away' => 'Perigord',
'scoreH' => '0',
'scoreA' => '0',
),
),
),
),
)