试图获得不显示的结果

Question

我只是想让一些基本的东西暂时出现，但没有用。它应该在屏幕上显示 1。我的逻辑错了吗？我将语句粘贴到控制台并得到 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'];

Answer 1

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

在上述方法中将连接作为第一个参数传递。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

Answer 2

首先你必须创建一个连接

<?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小写

试图获得不显示的结果

Trying to get a result not displaying

html

php

mysqli

echo