如果數據顯示在我的數據庫中,我想在我的複選框上輸出checked
屬性。測試數據庫結果以將複選框設置爲選中狀態
我試着做in_array()
但它並沒有因爲價值觀的工作(見資料庫)
我不知道我應該如何輸出他們很好
PHP代碼:
$general_list = "";
if($row['f_general'] == 'Bar, '){
$general_list .= "<label class=\"bg-danger\"><input type=\"radio\" name=\"general[]\" value=\"Bar, \" checked=\"checked\"/>Bar</label>";
}else{
$general_list .= "<label><input type=\"radio\" name=\"general[]\" value=\"Bar, \"/>Bar,</label>";
}
if($row['f_general'] == 'Restaurant,'){
$general_list .= "<label><input type=\"radio\" name=\"general[]\" value=\"Restaurant, \" checked=\"checked\"/>Restaurant,</label>";
}else{
$general_list .= "<label><input type=\"radio\" name=\"general[]\" value=\"Restaurant, \"/>Restaurant,</label>";
}
if($row['f_general'] == 'Coffee Shop,'){
$general_list .= "<label><input type=\"radio\" name=\"general[]\" value=\"Coffee Shop, \" checked=\"checked\"/>Coffee Shop,</label>";
}else{
$general_list .= "<label><input type=\"radio\" name=\"general[]\" value=\"Coffee Shop, \"/>Coffee Shop,</label>";
}
我的數據庫值是這樣保存的:
Bar, Restaurant, Coffee Shop, Concierge,
HTML CO DE:
<label><input type="checkbox" name="general[]" value="Bar, "/>Bar,</label>
<label><input type="checkbox" name="general[]" value="Restaurant, "/>Restaurant,</label>
<label><input type="checkbox" name="general[]" value="Coffee Shop, "/>Coffee Shop,</label>
<label><input type="checkbox" name="general[]" value="Concierge, "/>Concierge,</label>
<label><input type="checkbox" name="general[]" value="Business Center, "/>Business Center,</label>
<label><input type="checkbox" name="general[]" value="Salon, "/>Salon,</label>
感謝關於應該如何完成的想法。終於明白了 – Jonathan 2014-10-10 09:29:59