我有一个功能 get_me()。 get_me()返回页面的slug。
我还有mysql表表,其中包含以下数据,分别为 slug 和 block_to_echo :
----------------------------------------
| slug | block_to_echo |
----------------------------------------
| home | <b>Welcome</b> to home! |
| about | You're in about page! |
| services | Services, <i>yes</i>! |
----------------------------------------
现在,根据get_me(),我想得到block_to_echo的值,是的,回显它。
例如,如果 get_me()返回服务,我想回复&#34;服务,是!&#34;
已经定义了get_me()的功能。我只需要知道如何根据get_me()来获取block_to_echo。
显然,我不知道如何对此进行编码,所以如果我无法提供代码,我很抱歉。
答案 0 :(得分:0)
<?php
$database_host='database host goes here. (use localhost if on the same machine)';
$username='database username goes here';
$password='database password goes here';
$database_name='enter the name of your database here';
$database=mysqli_connect($database_host,$username,$password,$database_name)or die("Error " . mysqli_error($database));
$query='SELECT block_to_echo from table where slug='."'".get_me()."'";
$result=mysqli_query($database,$query);
if(!$result){
//Error handling code here for issue with query
}else{
//Everything went ok
if(mysqli_num_rows($result) < 1){
//There are no results for the value returned from get_me();
}else{
//We know we have at least one row. I'll assume it will only have a single row
$row=mysqli_fetch_assoc($result);
echo $row['block_to_echo']; //This should print the block to echo on the screen
}
}
?>