Php 根据函数获取mysql表列的值
我有一个函数,求你了。get_me返回页面的slug 我还有mysql表,其中包含以下数据,分别有slug列和block\u to\u echo列:Php 根据函数获取mysql表列的值,php,html,mysql,Php,Html,Mysql,我有一个函数,求你了。get_me返回页面的slug 我还有mysql表,其中包含以下数据,分别有slug列和block\u to\u echo列: ---------------------------------------- | slug | block_to_echo | ---------------------------------------- | home | <b>Welcome</b> to home! |
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| slug | block_to_echo |
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| home | <b>Welcome</b> to home! |
| about | You're in about page! |
| services | Services, <i>yes</i>! |
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现在,根据get_me,我想得到block_to_echo的值,是的,echo它
例如,如果get_me返回服务,我想回显服务,是的
已经定义了get_me的功能。我只需要知道如何根据get_me获得block_to_echo
显然,我不知道如何编写此代码,因此如果我无法提供代码,我很抱歉。但您知道其中一些,对吗?您的意思是$typ=get\me;mysqli_查询$dbh,从表'where slug='$typ';中选择block_to_echo然后获取并打印结果?是@flaschenpost。。。。。类似这样的东西…给了我这个错误:警告:mysqli\u query期望参数1是mysqli,string打开错误报告,并通过放置error\u reportingE\u ALL确保您已成功连接到mysql数据库;在打开php标记之后。如果它无法连接,你应该会出错。。连接到数据库没有问题。我只收到mysqli_查询错误..对我发布的代码进行了更新。复制并粘贴它完全符合我的要求,只需填写数据库用户名、密码等,以匹配您的密码是的,这有效。。。谢谢你的耐心
<?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
}
}
?>