从mysql数据库回显特定的行 [英] Echo specific rows from a mysql database

问题描述

我创建了一个表格，要求用户添加:名，姓，位置，状态

I have created a form that ask the user to add: first_name, last_name,location,status

一段时间后，我收到了5个输入.mysql表被命名为users，表数据如下:

After some time i have received 5 inputs. The mysql table is named users and the table data are like below:

id first_name location status  
== ========== ======== ========  
1   Chris       UK       Married  
2   Anton       Spain    Single  
3   Jacob       UK       Single  
4   Mike        Greece   Married  
5   George      UK       Married  

我也是通过POST方法接收输入的方式.所以:

Also the way i am receiving the inputs its via POST method. So:

$firstname=$_POST['FIRST_NAME']; // First Name: <input type="text" name="FIRST_NAME">
$location=$_POST['LOCATION']; // Location: <input type="text" name="LOCATION">
$status=$_POST['STATUS']; // Status: <input type="text" name="STATUS">

我创建了一个查询，以选择来自英国的所有已婚用户:

I created a query to select all users from UK that are married:

$query = "SELECT * FROM users WHERE location='UK' AND status='Married'";
$result = mysqli_query($dbc,$query); //$dbc is the connection to my database

$row = mysqli_fetch_array($results, MYSQLI_BOTH);

换句话说:

id first_name location status  
== ========== ======== ========  
1   Chris       UK       Married  
5   George      UK       Married

问题:

1)$ row数组是否如下图所示:

$row= array(
array(1 Chris UK Married),
array(5 George UK Married) 
);

2)在实施WHERE location ='UK'AND status ='Married'过滤之后，如何回显数据库的内容?

我需要这样:

Hello Chris! You are from UK and you are married!

Hello George! You are from UK and you are married!

我知道我必须使用foreach循环(回声数组)，但是我已经尝试过了，但是它不起作用.我尝试过的事情之一是在php.net中找到的东西:

I know that i have to use foreach loop (echo arrays) but i have tried it and it doesnt work.One of the things i tried was something i found in php.net:

使用list()打开嵌套数组的包装

Unpacking nested arrays with list()

(PHP 5> = 5.5.0)

(PHP 5 >= 5.5.0)

PHP 5.5通过提供list()作为值，增加了对数组数组进行迭代并将嵌套数组解压缩为循环变量的功能.

PHP 5.5 added the ability to iterate over an array of arrays and unpack the nested array into loop variables by providing a list() as the value.

例如:

<?php
$array = [
[1, 2],
[3, 4],
];

foreach ($array as list($a, $b)) {
// $a contains the first element of the nested array,
// and $b contains the second element.
echo "A: $a; B: $b\n";
}
?> 

当我使用上面的代码时，我收到以下错误:

When i use the above i am receiving the error below:

Parse error: syntax error, unexpected T_LIST in C:\wamp\www....

有什么建议吗?

据我了解，我必须以某种方式将ID与其他数据相关联.

As i understand, i somehow have to link the ID with the other data..

谢谢.

推荐答案

您可以使用类似这样的东西:

You may use something like this:

$query = "SELECT * FROM users WHERE location='UK' AND status='Married'";
$result = mysqli_query($dbc,$query);

while($row = mysqli_fetch_array($result, MYSQLI_BOTH)){
    printf("Hello %s! You are from %s and you are %s!\n", $row['first_name'], $row['location'],$row['status']);
}

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