Question

代码不言自明

如何将这两个SQL查询合并为一个？

类似“JOIN..ON ..或shop_tags.parent_tag_id = shop_tags_link.tag_id”

// query PARENT tags with this Design ID
$list_tags_link_query = "SELECT * FROM shop_tags_link JOIN shop_tags ON shop_tags_link.tag_id = shop_tags.tag_id WHERE shop_tags_link.design = '$design'";
$list_tags_link_exec = mysql_query($list_tags_link_query) or die(mysql_error());
    while($row = mysql_fetch_array($list_tags_link_exec)) {
    $tag_id = $row["tag_id"];
    $tag_fr = utf8_decode($row["tag_fr"]);
    $tag_en = utf8_decode($row["tag_en"]);
    echo "[<a href=\"$canon/?delete=$tag_id\">x</a>] $tag_fr ($tag_en), ";
        // query CHILD tags with the corresponding PARENT's tag_id
        $list_tags_child_query = "SELECT * FROM shop_tags WHERE parent_tag_id = '$tag_id'";
        $list_tags_child_exec = mysql_query($list_tags_child_query) or die(mysql_error());
        while($row = mysql_fetch_array($list_tags_child_exec)) {
            $tag_fr = utf8_decode($row["tag_fr"]);
            $tag_en = utf8_decode($row["tag_en"]);
            $tag_id = $row["tag_id"];
            echo "<font size=\"1\">[<a href=\"$canon/?delete=$tag_id\">x</a>] $tag_fr ($tag_en), </font>";
        }
    }

我只想要1个查询而不是2个

Answer 1

如果您只想使用一个查询，那么您可以尝试以下解决方案。但如果您的数据库表包含更多记录，则此查询需要更多时间来执行。

$list_tags_link_query = "SELECT stl.*,st.*,(select GROUP_CONCAT(CONCAT_WS('@@',in_t.tag_id,if(in_t.tag_fr <> '',in_t.tag_fr,''),if(in_t.tag_en <> '',in_t.tag_en,''))  SEPARATOR '|||') from shop_tags as in_t where in_t.parent_tag_id = stl.tag_id) as childs_details FROM shop_tags_link stl JOIN shop_tags st ON stl.tag_id = st.tag_id WHERE stl.design = '$design'";
$list_tags_link_exec = mysql_query($list_tags_link_query) or die(mysql_error());
while ($row = mysql_fetch_array($list_tags_link_exec)) {
    $tag_id = $row["tag_id"];
    $tag_fr = utf8_decode($row["tag_fr"]);
    $tag_en = utf8_decode($row["tag_en"]);
    echo "[<a href=\"$canon/?delete=$tag_id\">x</a>] $tag_fr ($tag_en), ";

    $childs_details=$row["childs_details"];
    if($childs_details!=""){
        $child_row=explode("|||",$childs_details);
        if(is_array($child_row) && count($child_row)>0){
            for($i=0;$i<count($child_row);$i++){
                $child_row_detail=explode("@@",$child_row[$i]);

                $tag_id = $child_row_detail[0];
                $tag_fr = utf8_decode($child_row_detail[1]);
                $tag_en = utf8_decode($child_row_detail[2]);
                echo "<font size=\"1\">[<a href=\"$canon/?delete=$tag_id\">x</a>] $tag_fr ($tag_en), </font>";
            }
        }
    }


}

Answer 2

我认为你几乎是正确的代码，但你犯了一个错误。 $ row 在循环中被覆盖所以只需更正它。请检查下面的代码。

// query PARENT tags with this Design ID
$list_tags_link_query = "SELECT * FROM shop_tags_link JOIN shop_tags ON shop_tags_link.tag_id = shop_tags.tag_id WHERE shop_tags_link.design = '$design'";
$list_tags_link_exec = mysql_query($list_tags_link_query) or die(mysql_error());
    while($row = mysql_fetch_array($list_tags_link_exec)) {
    $tag_id = $row["tag_id"];
    $tag_fr = utf8_decode($row["tag_fr"]);
    $tag_en = utf8_decode($row["tag_en"]);
    echo "[<a href=\"$canon/?delete=$tag_id\">x</a>] $tag_fr ($tag_en), ";
        // query CHILD tags with the corresponding PARENT's tag_id
        $list_tags_child_query = "SELECT * FROM shop_tags WHERE parent_tag_id = '$tag_id'";
        $list_tags_child_exec = mysql_query($list_tags_child_query) or die(mysql_error());
        while($child_row = mysql_fetch_array($list_tags_child_exec)) {
            $tag_fr = utf8_decode($child_row["tag_fr"]);
            $tag_en = utf8_decode($child_row["tag_en"]);
            $tag_id = $child_row["tag_id"];
            echo "<font size=\"1\">[<a href=\"$canon/?delete=$tag_id\">x</a>] $tag_fr ($tag_en), </font>";
        }
    }

使用JOIN + OR语句将2个mysql查询合并为一个

2 个答案: