我有一些PHP变量集,我正在从中创建一个多维数组。现在,在该数组中,我想检查重复项的特定键([font]
)。
如果找到重复项,则[lang]
和[weight]
的相应值和相应值应合并。
这是我到目前为止所尝试的内容(这会取消/删除数组中的重复值):
// Font [0]
$font_1 = "Poppins";
$font_1_l = "Hindi, English";
$font_1_w = "700, 700i";
// Font [1]
$font_2 = "Lora";
$font_2_l = "Vietnamese, Japanese";
$font_2_w = "200, 300, 400, 400i";
// Font [2]
$font_3 = "Noto Sans";
$font_3_l = "Punjabi, Latin, Hindi";
$font_3_w = "200, 200i, 300, 300i, 400, 500";
// Font [3]
$font_4 = "Lora";
$font_4_l = "Greek, Roman, Vietnamese";
$font_4_w = "400, 400i, 500, 500b";
// Array of all the values
$font_f = array( array( 'font' => $font_1, 'lang' => $font_1_l, 'weight' => $font_1_w ), array( 'font' => $font_2, 'lang' => $font_2_l, 'weight' => $font_2_w ), array( 'font' => $font_3, 'lang' => $font_3_l, 'weight' => $font_3_w ), array( 'font' => $font_4, 'lang' => $font_4_l, 'weight' => $font_4_w ) );
// Printing the array for testing
echo "<pre>";
print_r( array_map("unserialize", array_unique(array_map("serialize", $font_f))) );
// Removing duplicates
$font_f_copy = $font_f; // Copy of $font_f for modification
$fonts = array(); // To get unique fonts
for( $i=0; $i<count($font_f); $i++ ) {
if ( in_array( $font_f[$i]['font'], $fonts ) ) {
unset($font_f_copy[$i]);
}
else {
$fonts[] = $font_f[$i]['font'];
}
}
// Printing $font_f_copy for testing
print_r($font_f_copy);
输出:
Array
(
[0] => Array
(
[font] => Poppins
[lang] => Hindi, English
[weight] => 700, 700i
)
[1] => Array
(
[font] => Lora
[lang] => Vietnamese, Japanese
[weight] => 200, 300, 400, 400i
)
[2] => Array
(
[font] => Noto Sans
[lang] => Punjabi, Latin, Hindi
[weight] => 200, 200i, 300, 300i, 400, 500
)
[3] => Array
(
[font] => Lora
[lang] => Greek, Roman, Vietnamese
[weight] => 400, 400i, 500, 500b
)
)
Array
(
[0] => Array
(
[font] => Poppins
[lang] => Hindi, English
[weight] => 700, 700i
)
[1] => Array
(
[font] => Lora
[lang] => Vietnamese, Japanese
[weight] => 200, 300, 400, 400i
)
[2] => Array
(
[font] => Noto Sans
[lang] => Punjabi, Latin, Hindi
[weight] => 200, 200i, 300, 300i, 400, 500
)
)
正如您在上面的代码中看到的,Font [1]和Font [3]将具有[font]
即Lora的相同值,因此Font的[lang]
和[weight]
[1]应分别与{[1}}和[lang]
合并为Font [3]。
我想知道如何继续实现这一目标。
答案 0 :(得分:1)
而不是for循环,我个人会做类似的事情:
$result = [];
foreach ($font_f as $f) {
if (isset($result[$f["font"]])) {
$result[$f["font"]] = [
"font" => $f["font"],
"lang" => $f["lang"]." ,".$result[$f["font"]]["lang"],
"weight" => $f["weight"]." ,".$result[$f["font"]]["weight"]
];
} else {
$result[$f["font"]] = $f;
}
}
答案 1 :(得分:1)
从$array=[];
foreach($font_f as $a){
$array[$a["font"]]["font"]=$a["font"]; // declare or lazy overwrite if duplicate
foreach(array("lang","weight") as $col){ // cycle through the concat-able elements
if(isset($array[$a["font"]][$col])){
$temp_str=$array[$a["font"]][$col].", ".$a[$col]; // concat
}else{
$temp_str=$a[$col]; // declare
}
$temp_arr=array_unique(explode(', ',$temp_str)); // split & remove duplicates
sort($temp_arr); // sort
$array[$a["font"]][$col]=implode(', ',$temp_arr); // rejoin
}
}
ksort($array); // sort subarrays by font name
$array=array_values($array); // replace temporary associative keys with indices
echo "<pre>";
print_r($array);
echo "</pre>";
开始,这将:
以下是代码:
Array(
[0] => Array(
[font] => Lora
[lang] => Greek, Japanese, Roman, Vietnamese
[weight] => 200, 300, 400, 400i, 500, 500b
),
[1] => Array(
[font] => Noto Sans
[lang] => Hindi, Latin, Punjabi
[weight] => 200, 200i, 300, 300i, 400, 500
),
[2] => Array(
[font] => Poppins
[lang] => English, Hindi
[weight] => 700, 700i
)
)
输出:
{{1}}