PHP 帮助构建查询 [英] PHP help with building query

问题描述

我有几个下拉列表，如果没有选择任何选项，则值为 = ""...

I have several drop lists where if no option is selected then the value is = ""...

我不知道如何在 PHP 中为 mysql 构建查询.

I cant figure out how to build the query for mysql in PHP.

查询 = SELECT * FROM db

query = SELECT * FROM db

推荐答案

我假设您有这样的选择:

I assume you have a select like this:

<select name="data[]" multiple="multiple">
<option value="A">A</option>
<option value="B">B</option>
<option value="C">C</option>
</select>

您的 php 可能类似于

Your php can be something like

<?php
$data = array();
$data = $_POST['data'];
$query = "select * from table";
if (count($data > 0)) {
    for ($i = 0; $i < count($data); $i++) {
        $data[$i] = "'{$data[$i]}'";
    }
    $query .= " where field in (".implode(",", $data).")";
}

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