尝试不显示结果

时间:2019-03-04 18:04:23

标签: php html mysqli echo

我只是想让一些基本的东西现在出现,并且不起作用。它应该在屏幕上显示1。我的逻辑错了吗?我将声明粘贴到控制台中并得到1。

<?php

$sql = mysqli_query("Select moist_measure_avail from sigh_in_account where moist_measure_avail = '1'");

$result = mysqli_fetch_array($sql);

echo $result['moist_measure_avail'];

2 个答案:

答案 0 :(得分:0)

$sql = mysqli_query("Select moist_measure_avail from sigh_in_account where moist_measure_avail = '1'");

将连接作为上述方法中的第一个参数。像这样

$sql = mysqli_query($conn,"Select moist_measure_avail from sigh_in_account where moist_measure_avail = '1'");

其中$ conn是mysql连接

答案 1 :(得分:0)

首先您必须创建一个连接

<?php
    $conn = mysqli_connect("localhost "," root","","dbname");
?>

然后您必须在查询中包含conn变量

$sql = mysqli_query( $conn, " SELECT moist_measure_avail from sigh_in_account WHERE moist_measure_avail = '1'");

sql语句必须全部大写或全部全部小写

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