Php mysql中的If子句select

我的全部问题是：

$sql = "SELECT vehicle_id,  fuel_type, gearbox_type, door_type, engine_package_name, FORMAT(engine_cap, 1)engine_cap, engine_power, FORMAT(fuel_cons, 1)fuel_cons, FORMAT(std_price, 0)std_price 
FROM vehicle 
WHERE model_id = '$model_id' 
AND fuel_type IN ('$fuel_type1', '$fuel_type2', '$fuel_type3', '$fuel_type4') AND gearbox_type IN( '$gearbox_type1', '$gearbox_type2', '$gearbox_type3', '$gearbox_type4') AND door_type IN ('$door_type1', '$door_type2', '$door_type3', '$door_type4')" ;

如果$fuel_type1等于零，我想跳过下面显示的查询部分

AND fuel_type IN ('$fuel_type1', '$fuel_type2', '$fuel_type3', '$fuel_type4') 

我试过了

$sql = "SELECT vehicle_id,  fuel_type, gearbox_type, door_type, engine_package_name, FORMAT(engine_cap, 1)engine_cap, engine_power, FORMAT(fuel_cons, 1)fuel_cons, FORMAT(std_price, 0)std_price 
FROM vehicle 
WHERE model_id = '$model_id' ";

if($fuel_type1 != 0) 
{ 
    $sql += “ AND fuel_type IN ('$fuel_type1', '$fuel_type2', '$fuel_type3', '$fuel_type4')”;
}

$sql = "SELECT vehicle_id, fuel_type, gearbox_type, door_type, engine_package_name, FORMAT(engine_cap, 1)engine_cap, engine_power, FORMAT(fuel_cons, 1)fuel_cons, FORMAT(std_price, 0)std_price 
FROM vehicle 
WHERE model_id='$model_id’ 
AND IF($fuel_type1 <> 0,'fuel_type','0') IN ('$fuel_type1', '$fuel_type2', '$fuel_type3', '$fuel_type4’)";

没用

我也试过了

$sql = "SELECT vehicle_id,  fuel_type, gearbox_type, door_type, engine_package_name, FORMAT(engine_cap, 1)engine_cap, engine_power, FORMAT(fuel_cons, 1)fuel_cons, FORMAT(std_price, 0)std_price 
FROM vehicle 
WHERE model_id = '$model_id' ";

if($fuel_type1 != 0) 
{ 
    $sql += “ AND fuel_type IN ('$fuel_type1', '$fuel_type2', '$fuel_type3', '$fuel_type4')”;
}

$sql = "SELECT vehicle_id, fuel_type, gearbox_type, door_type, engine_package_name, FORMAT(engine_cap, 1)engine_cap, engine_power, FORMAT(fuel_cons, 1)fuel_cons, FORMAT(std_price, 0)std_price 
FROM vehicle 
WHERE model_id='$model_id’ 
AND IF($fuel_type1 <> 0,'fuel_type','0') IN ('$fuel_type1', '$fuel_type2', '$fuel_type3', '$fuel_type4’)";

不起作用：

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任何帮助都将不胜感激。

问题在于连接过程中

在PHP中，应该有b a do t。连接非+符号

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第一次尝试应该可以，您收到了什么错误？问题是在PHP的连接过程中应该有b a t。连接not+符号“``if$fuel\u type1！=0{$sql.=“和燃料类型在“$fuel_type1”、“$fuel_type2”、“$fuel_type3”、“$fuel_type4”中；}```