从mysql数据库中回显特定的行

时间:2014-01-15 13:27:55

标签: php mysql sql arrays

我创建了一个要求用户添加的表单: first_name,last_name,location,status

一段时间后,我收到了5个输入。 mysql表名为users,表数据如下:

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  

另外我通过POST方式接收输入的方式。所以:

$firstname=$_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">

我创建了一个查询来选择来自英国的所有已婚用户:

$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数组是否如下所示:

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

2)如何在实施过滤后回显数据库的内容WHERE location ='UK'AND status ='Married'?

我需要它是这样的:

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(echo数组),但我已经尝试了它并且它不起作用。我试过的一些东西是我在php.net中找到的东西:

使用list()

解压缩嵌套数组

(PHP 5> = 5.5.0)

PHP 5.5增加了迭代数组数组的能力,并通过提供list()作为值将嵌套数组解压缩为循环变量。

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";
}
?> 

当我使用上述内容时,我收到以下错误:

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

有什么建议吗?

据我了解,我不得不将ID与其他数据联系起来..

提前致谢。

3 个答案:

答案 0 :(得分:3)

您可以使用以下内容:

$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']);
}

答案 1 :(得分:1)

mysqli_fetch_array返回一行。你需要循环来获取所有行。您可以从行中获取值并将其放在以下字符串中:

while($row = mysqli_fetch_array($results, MYSQLI_BOTH)){
    echo "Hi ".$row['first_name'] . " You are from ".$row['location']." and you are ".$row['status']."!";
}

答案 2 :(得分:0)

问题2:您可以在sql查询中添加格式我猜:

"SELECT OUTPUT = "Hello" + name + "! You are from UK and you are married!" FROM users WHERE location='UK' AND status='Married'";