有没有办法将数组转换成这样:
Array ( [level-1] => 2 [quarter-1] => 1 [year-1] => 2014 [level-2] => 2 [quarter-2] => 2 [year-2] => 2015 [level-49] => 2 [quarter-49] => 2 [year-49] => 2015 [level-58] => 1 [quarter-58] => 1 [year-58]并且只接受关键字后面的数字,以便在sql表中使用它们。例如,该表将如下所示:
ID 1 Level 2 Quarter 1 Year 2014
ID 2 level 2 quater 2 Year 2015
ID 49 level 2 quarter 2 year 2015 等
我试过了
if(!empty($_POST)){
print_r ($_POST);
echo "<br/><br/>";
$s=$_POST;
echo $abc= implode(',', $s);
for($a=0;$a<count($s);$a++){
$ar=explode(',',$abc);
echo $var=$ar[$a];
}
}但我得到的结果是:
2,1,2014,2,2,2015,2,2,2015...我还需要身份证来显示。但最重要的是,我不知道如何解释结果以将它们放入数据库中。
发布于 2013-03-27 17:10:01
试试这个:
$array = array ( "level-1" => 2, "quarter-1" => 1, "year-1" => 2014, "level-2" => 2, "quarter-2" => 2, "year-2" => 2015, "level-49" => 2, "quarter-49" => 2, "year-49" => 2015, "level-58" => 1, "quarter-58" => 1, "year-58"=>2016);
foreach(array_chunk($array,3,true) as $val){
foreach($val as $k=>$v){
if(strpos($k, "level") !== false){
$temp = explode("-",$k);
$id = $temp[1];
$level = $v;
}
if(strpos($k, "quarter") !== false){
$quarter = $v;
}
if(strpos($k, "year") !== false){
$year = $v;
}
}
echo "ID ".$id." Level ".$level." Quarter ".$quarter." Year ".$year;
echo "<br>";
}输出:
ID 1 Level 2 Quarter 1 Year 2014
ID 2 Level 2 Quarter 2 Year 2015
ID 49 Level 2 Quarter 2 Year 2015
ID 58 Level 1 Quarter 1 Year 2016发布于 2013-03-27 17:09:22
我不知道如何用PHP编写代码,所以我将向您解释这样做的逻辑。
for(int i=0;i<array.length;i=i+3) {
//get value from array[i]; get the level-x field and retrieve the x value, then extract integers accordingly from it after keywords. In java StringTokenizer can be used for this, or regex etc
//again get value from array[i+1] and extract the integer
//get the value from array[i+2] and do the same as above.
}
如果数组是随机排列的,则可以将其与level-x进行匹配,以获得要插入的确切元组。因此,level-x,x值充当对所有相关值进行分组的索引。
https://stackoverflow.com/questions/15654902
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