所以我在我的資料庫中創建了一個接受 JSON 陣列的列，在這些陣列中是 item id，我想在這些 id 上創建一個回圈并將其插入到 sql select 陳述句中以顯示在陣列中具有 id 的所有專案，但是我對如何做到這一點一無所知。
$sql = "SELECT bookmarks FROM `accounts` WHERE firstname='".$firstname."' && lastname='".$lastname."' ";
$sth = mysqli_query($conn, $sql);
$rows = array();
while($r = mysqli_fetch_assoc($sth)) {
$rows[] = $r;
}
foreach($rows as $key ){ ?????
$bookmark = "SELECT * FROM books id='".$key."'";
$bookselect = mysqli_query($conn, $bookmark);
while($row = mysqli_fetch_assoc($bookselect)) {
}
}????
uj5u.com熱心網友回復：
當您將陣列存盤在資料庫中時，您需要先將其轉換為陣列和 ID：
例如檢查下面的代碼：
$sql = "SELECT bookmarks FROM `accounts` WHERE firstname='".$firstname."' && lastname='".$lastname."' ";
$sth = mysqli_query($conn, $sql);
$rows = array();
while($r = mysqli_fetch_assoc($sth)) {
$rd = json_decode($r['bookmarks'], true);
foreach($rd as $ids){
$rows[] = $ids;
}
}
$bookmark = "SELECT * FROM books WHERE id IN('".implode("','",$rows)."'";
$bookselect = mysqli_query($conn, $bookmark);
while($row = mysqli_fetch_assoc($bookselect)) {
print_r($row);
}
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