从函数内部调用mysqli_query时不起作用[复制]

This question already has an answer here:

这个问题在这里已有答案:

Reference: What is variable scope, which variables are accessible from where and what are “undefined variable” errors? 3 answers

参考:什么是变量范围,哪些变量可以从哪里访问,什么是“未定义变量”错误? 3个答案

Here is my code:

这是我的代码:

<?php
$db_host        = 'localhost';
$db_user        = 'root';
$db_pass        = 'root';
$db_database    = 'drmahima_com';
$link = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
mysqli_query($link, "SET names UTF8");
function temp() {
    $patient = mysqli_fetch_assoc(mysqli_query($link, "SELECT name,dob FROM patients WHERE id='69'"));
    echo $patient['name'];
}
temp();
?>

When I run this, the query doesn't seem to execute and nothing shows up on the page. However, if I remove the function and just change the code to this:

当我运行它时,查询似乎没有执行,并且页面上没有显示任何内容。但是,如果我删除该函数,只需将代码更改为:

<?php
$db_host        = 'localhost';
$db_user        = 'root';
$db_pass        = 'root';
$db_database    = 'drmahima_com';
$link = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
mysqli_query($link, "SET names UTF8");
// function temp() {
    $patient = mysqli_fetch_assoc(mysqli_query($link, "SELECT name,dob FROM patients WHERE id='69'"));
    echo $patient['name'];
// }
// temp();
?>

It works, and the patient's name is displayed.

它有效,并显示患者的姓名。

What is going on here?

这里发生了什么?

2 个解决方案

#1

Pass the $link to function as a parameter. Try this:

将$ link作为参数传递给函数。试试这个:

<?php
$db_host        = 'localhost';
$db_user        = 'root';
$db_pass        = 'root';
$db_database    = 'drmahima_com';
$link = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
mysqli_query($link, "SET names UTF8");
function temp($link) {
    $patient = mysqli_fetch_assoc(mysqli_query($link, "SELECT name,dob FROM patients WHERE id='69'"));
    echo $patient['name'];
}
temp($link);
?>

#2

Pass $link to the function. It is not working because of the scope of that variable. Try with -

将$ link传递给该函数。由于该变量的范围,它无法正常工作。尝试 -

function temp($link) {
    $patient = mysqli_fetch_assoc(mysqli_query($link, "SELECT name,dob FROM patients WHERE id='69'"));
    echo $patient['name'];
}
temp($link);

#1