更新----
使用此方法后:
foreach($_POST as $k => $v) {
$params[] = empty($v)? "NULL":$v;
}
$params_string_values = "'" . implode("','",$params) . "'";
$param_name_list = "tu_id,tu_status,tu_name,tu_fk_tt_id,tu_mon_1_s,tu_mon_1_e,tu_mon_2_s,tu_mon_2_e,tu_mon_3_s,tu_mon_3_e,tu_tue_1_s,tu_tue_1_e,tu_tue_2_s,tu_tue_2_e,tu_tue_3_s,tu_tue_3_e,tu_wed_1_s,tu_wed_1_e,tu_wed_2_s,tu_wed_2_e,tu_wed_3_s,tu_wed_3_e,tu_thu_1_s,tu_thu_1_e,tu_thu_2_s,tu_thu_2_e,tu_thu_3_s,tu_thu_3_e,tu_fri_1_s,tu_fri_1_e,tu_fri_2_s,tu_fri_2_e,tu_fri_3_s,tu_fri_3_e,tu_sat_1_s,tu_sat_1_e,tu_sat_2_s,tu_sat_2_e,tu_sat_3_s,tu_sat_3_e,tu_sun_1_s,tu_sun_1_e,tu_sun_2_s,tu_sun_2_e,tu_sun_3_s,tu_sun_3_e";
$param_values = "'','1',{$params_string_values}";
$insert_query = mysql_query("INSERT into turn_conf( {$param_name_list} ) values ({$param_values})");
它创建了一个看起来有效的查询(我在这里粘贴了它),但是没有 NULL 值存储在数据库中,所有 NULL 值都以“00:00”的形式进入数据库:
INSERT into turn_conf( tu_id,tu_status,tu_name,tu_fk_tt_id,tu_mon_1_s,tu_mon_1_e,tu_mon_2_s,tu_mon_2_e,tu_mon_3_s,tu_mon_3_e,tu_tue_1_s,tu_tue_1_e,tu_tue_2_s,tu_tue_2_e,tu_tue_3_s,tu_tue_3_e,tu_wed_1_s,tu_wed_1_e,tu_wed_2_s,tu_wed_2_e,tu_wed_3_s,tu_wed_3_e,tu_thu_1_s,tu_thu_1_e,tu_thu_2_s,tu_thu_2_e,tu_thu_3_s,tu_thu_3_e,tu_fri_1_s,tu_fri_1_e,tu_fri_2_s,tu_fri_2_e,tu_fri_3_s,tu_fri_3_e,tu_sat_1_s,tu_sat_1_e,tu_sat_2_s,tu_sat_2_e,tu_sat_3_s,tu_sat_3_e,tu_sun_1_s,tu_sun_1_e,tu_sun_2_s,tu_sun_2_e,tu_sun_3_s,tu_sun_3_e ) values ('','1','12345555','1','10:00','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL','NULL')
这是它在 DB 上记录的内容:
我有这个查询,其中有时变量 '$va1'
、'$val2'
和 '$val3'
没有值:
$insert = mysql_query("INSERT INTO turn_conf (tu_id,value1,value2,value3) VALUES ('','$va1','$val2','$val3')") or die (mysql_error());
如果这些变量中的任何一个都没有存储值,则必须将与其相关的任何内容发送到数据库(以便在数据库上存储 NULL 值),例如如果只有 '$val1'
存储信息,最终查询必须是:
$insert = mysql_query("INSERT INTO turn_conf (tu_id,value1) VALUES ('','$va1')") or die (mysql_error());
为了解决这个问题,我为每个变量创建了一个结构,检查它是否存储了一些东西,如果没有,就什么都不声明也不发送:
if ($_POST['value1'] == ""){
$val1_p = "";
$val1_s = "";
}else {
$val1_p = ",value1";
$val1_sv = $_POST['value1'];
$val1_s = ", '$val1_sv'" ;}
if ($_POST['value2'] == ""){
$val2_p = "";
$val2_s = "";
}else {
$val2_p = ",value2";
$val2_sv = $_POST['value2'];
$val2_s = ", '$val2_sv'" ;}
if ($_POST['value3'] == ""){
$val3_p = "";
$val3_s = "";
}else {
$val3_p = ",value3";
$val3_sv = $_POST['value3'];
$val3_s = ", '$val3_sv'" ;}
和:
$insert = mysql_query("INSERT INTO turn_conf (tu_id $val1_p $val2_p $val3_p) VALUES ('' $val1_s $val2_s $val3_s)") or die (mysql_error());
这行得通,并创建了正确的查询,但我想知道您是否觉得这个方法合适,或者是否选择另一个更有效的方法会更好。请不要在这个例子中我只使用了 3 个变量,但是这个真实的查询有 43 个变量,我因为数据量而提出这个问题。
最佳答案
$insert = mysql_query("INSERT INTO turn_conf (tu_id,value1,value2,value3) VALUES ('','".((isset($va1))?"'".$va1."'":"NULL")."','".((isset($va2))?"'".$va2."'":"NULL")."','".((isset($va3))?"'".$va3."'":"NULL")."')") or die (mysql_error());
基本上您想要测试该值是否已设置。我们为此使用简短的 if 符号:
isset($va1)?"'".$va1."'":"NULL"
如果设置了 $va1(有值),我们将在查询中放入“'value'”,否则为空值的“NULL”。
如果你也想测试一个空字符串:
(isset($va1) && $va1 != '')?"'".$va1."'":"NULL"
关于php - 在 DB 上插入 NULL 值(方法),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23490883/