Question

当这些元素显示在菜单上时，我想反转它们的顺序：

function buscarSubCateg($categ_id) {

    global $con;

    $buscar_subcateg = "SELECT * FROM subcateg WHERE categ_id = '$categ_id'";

    $run_subcateg = mysqli_query($con, $buscar_subcateg);

    $i=0;
    while ($row_subcateg = mysqli_fetch_array($run_subcateg)) {

        $subcateg_id = $row_subcateg['subcateg_id'];
        $subcateg_name = $row_subcateg['subcateg_name'];

        echo '<div class="sectionSub"><a class="linkSubCatergoria" href="#">'.$subcateg_name.'</a></div>'.buscarTipos($subcateg_id);
        $i++;
    }
}

function buscarTipos($subcateg_id) {

    global $con;

    $buscar_tipo = "SELECT * FROM tipoprod WHERE subcateg_id = '$subcateg_id'";

    $run_tipo = mysqli_query($con, $buscar_tipo);

    while ($row_tipo = mysqli_fetch_array($run_tipo)) {

        $tipo_id = $row_tipo['tipoprod_id'];
        $tipo_name = $row_tipo['tipoprod_name'];

        echo "<div class='sectionTipo'><a class='linkTipo' href='#'>$tipo_name</a></div>";

    }
}

在此回显中：

echo '<div class="sectionSub"><a class="linkSubCatergoria" href="#">'.$subcateg_name.'</a></div>'.buscarTipos($subcateg_id);`

我希望函数buscarTipos()的结果显示在$subcateg_name之后。现在的做法与此相反，首先显示buscarTipos()的结果，然后显示$subcateg_name。

谢谢。

Answer 1

用以下代码替换当前代码：

function buscarSubCateg($categ_id) {

    global $con;

    $buscar_subcateg = "SELECT * FROM subcateg WHERE categ_id = '$categ_id'";

    $run_subcateg = mysqli_query($con, $buscar_subcateg);

    $i=0;
    while ($row_subcateg = mysqli_fetch_array($run_subcateg)) {

        $subcateg_id = $row_subcateg['subcateg_id'];
        $subcateg_name = $row_subcateg['subcateg_name'];
        $tip = buscarTipos($subcateg_id);
        echo $tip;
        echo '<div class="sectionSub"><a class="linkSubCatergoria" href="#">'.$subcateg_name.'</a></div>';
        $i++;
    }
}

function buscarTipos($subcateg_id) {

    global $con;

    $buscar_tipo = "SELECT * FROM tipoprod WHERE subcateg_id = '$subcateg_id'";

    $run_tipo = mysqli_query($con, $buscar_tipo);
    $str = '';
    while ($row_tipo = mysqli_fetch_array($run_tipo)) {

        $tipo_id = $row_tipo['tipoprod_id'];
        $tipo_name = $row_tipo['tipoprod_name'];

        $str = $str."<div class='sectionTipo'><a class='linkTipo' href='#'>$tipo_name</a></div>";

    }
    return $str;
}

Answer 2

尝试更改第34行

echo "<div class='sectionTipo'><a class='linkTipo' href='#'>$tipo_name</a></div>";

进入

return "<div class='sectionTipo'><a class='linkTipo' href='#'>$tipo_name</a></div>";

然后，调用函数将在回显sectionSub之前回显sectionTipo

或将您的SQL更改为

$sql = 'select a.categ_id, a.subcateg_id, b.tipoprod_id, b.tipoprod_name
from subcateg as a, tipoprod as b where categ_id='$categ_id' 
and a.subcateg_id = b.subcateg_id';

并完全失去辅助功能。这样一来，MySQL便可以在两个表之间进行联接，并且易于使用一个数量级。

PHP中元素的显示顺序颠倒

2 个答案: