从所有表中选择行的简便方法 [英] easy ways to select rows from all table

问题描述

我想在我的mysql数据库中选择一些行，我有3个具有相同结构的表，我尝试使用我创建的脚本并且此方法有效，但是似乎我的脚本中的行太多了. /p>

i want to select some rows in my mysql databases, I have 3 tables with the same structure, I have tried using the script that I created and this works, but it seems like it takes too many lines in my script.

<?php
$check1 = mysqli_query($con, "SELECT * FROM text WHERE keyword='$string'");
$check2 = mysqli_query($con, "SELECT * FROM sticker WHERE keyword='$string'");
$check3 = mysqli_query($con, "SELECT * FROM image WHERE keyword='$string'");
if(mysqli_num_rows($check1) == 0 && mysqli_num_rows($check2) == 0 && mysqli_num_rows($check3) == 0) {
    echo "success";
}
?>

还有另一种方法可以缩短我的脚本吗?谢谢:)

is there another way to shorten my script above ? thank you :)

推荐答案

如果所有表都具有相同的结构，则可以使用UNION查询一次返回所有三个表中的行:

If all the tables have the same structure, you can use a UNION query to return rows from all three tables at once:

$check = mysqli_query($con, "SELECT * FROM text WHERE keyword='$string'
                             UNION
                             SELECT * FROM sticker WHERE keyword='$string'
                             UNION
                             SELECT * FROM image WHERE keyword='$string'");

这将为您提供与三个现有查询相同的行集.请注意，将无法确定给定行来自哪个表，因此您可能需要添加一个额外的字段来表示该行，例如

This will give you the equivalent set of rows that your three existing queries give you. Note that there will be no way to determine which table a given row came from, so you may want to add an extra field to indicate that e.g.

$check = mysqli_query($con, "SELECT *, 'text' AS src FROM text WHERE keyword='$string'
                             UNION
                             SELECT *, 'sticker' AS src FROM sticker WHERE keyword='$string'
                             UNION
                             SELECT *, 'image' AS src FROM image WHERE keyword='$string'");

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