我有一個數組的數組。還有另一個數組的數組。如果存在一個具有 common 值的數組，我希望將其從第一個數組中刪除：這是我的代碼function availableTables($tables, $tablesWithDate) { $tablesReturn = array(); foreach($tables as $table) { foreach($tablesWithDate as $twd) { if($table['table_no'] == $twd['number']){ echo "true"; } else { echo "false"; } } } return $tablesReturn; }這樣，除了 2 次返回 true 之外，每次迭代的輸出都是 false。這是正確的，但我想從第一個數組中刪除它們。我努力了：function availableTables($tables, $tablesWithDate) { $tablesReturn = array(); foreach($tables as $table) { foreach($tablesWithDate as $twd) { if($table['table_no'] == $twd['number']){ unset($table); } else { echo "false"; } } } return $tablesReturn; }但不起作用。它說未定義的變量表。我也嘗試過unset($tables[$table])$TablesWithDate：Array( [0] => Array ( [id] => 206 [number] => 150 [capacity] => 4 [booking_code] => qhJEHcWnzty062DD [reservation_date] => 2020-07-09 01:00:00 [start_time] => 12:30:00.000000 [end_time] => 14:15:00.000000 ) [1] => Array ( [id] => 206 [number] => 150 [capacity] => 4 [booking_code] => ym9dP1aZtFstP3WM [reservation_date] => 2020-07-22 01:00:00 [start_time] => 20:00:00.000000 [end_time] => 21:45:00.000000 ))表：( [0] => Array ( [id] => 159 [table_no] => 150 [capacity] => 6 [shape] => large_rectangle [joinable] => 1 [area] => 1 [baby_friendly] => 1 [premise_code] => LJJIDHhRN2ho1e3h [area_name] => Ferkin [premise_name] => An Poitin Stil )
1 回答
LEATH
TA貢獻1936條經驗 獲得超7個贊
使用表數組的索引（代碼中的$key）并使用 取消設置unset($tables[$key]。
function availableTables($tables, $tablesWithDate) {
$tablesReturn = array();
foreach($tables as $key => $table) {
foreach($tablesWithDate as $twd) {
if($table['table_no'] == $twd['number']){
unset($tables[$key]);
} else {
echo "false";
}
}
}
return $tables;
}
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