PHP条件查询来自GET方法

时间:2018-04-07 01:39:42

标签: php

我需要为我的PHP代码添加条件,为了加载JSON文件,我需要加载第一个查询1,如果 pos等于1 ,但是如果 pos不同于1 ,加载查询2

我该怎么做?

我的代码

        $pos = $_GET["pos"];

        if ($pos = "1"){
        // query 1
            $query = mysqli_query($con, "SELECT * FROM escolar WHERE r_ruta = '$r_ruta' and jornada = '$jornada' and pos = '$pos'");
        }else {
        // query 2
            $query = mysqli_query($con, "SELECT * FROM escolar WHERE r_ruta = '$r_ruta' and jornada = '$jornada' and pos < '$pos' order by pos desc limit 30");
        }

1 个答案:

答案 0 :(得分:-1)

你就在那儿。您必须使用==,因为如果没有,您只需使用=将$ pos设置为1。

   $pos = $_GET["pos"];

    if ($pos == "1"){
    // query 1
        $query = mysqli_query($con, "SELECT * FROM escolar WHERE r_ruta = '$r_ruta' and jornada = '$jornada' and pos = '$pos'");
    }else {
    // query 2
        $query = mysqli_query($con, "SELECT * FROM escolar WHERE r_ruta = '$r_ruta' and jornada = '$jornada' and pos < '$pos' order by pos desc limit 30");
    }