如何从此MySQL查询中获取PHP变量？

Question

我正在使用PHP / MySQL处理资产数据库问题。

在此脚本中，我想通过资产ID搜索我的资产，并使其返回所有相关字段。

首先，我查询数据库资产表并找到资产的类型。 然后根据类型运行3个查询中的1个。

<?php

//make database connect
mysql_connect("localhost", "asset_db", "asset_db") or die(mysql_error());
mysql_select_db("asset_db") or die(mysql_error());

//get type of asset
$type = mysql_query("
SELECT asset.type
From asset
WHERE asset.id = 93120
")
or die(mysql_error());

switch ($type){
    case "Server":
        //do some stuff that involves a mysql query
        mysql_query("
        SELECT asset.id
        ,asset.company
        ,asset.location
        ,asset.purchase_date
        ,asset.purchase_order
        ,asset.value
        ,asset.type
        ,asset.notes
        ,server.manufacturer
        ,server.model
        ,server.serial_number
        ,server.esc
        ,server.user
        ,server.prev_user
        ,server.warranty
        FROM asset
        LEFT JOIN server
        ON server.id = asset.id
        WHERE asset.id = 93120
        ");
        break;
    case "Laptop":
        //do some stuff that involves a mysql query
        mysql_query("
        SELECT asset.id
        ,asset.company
        ,asset.location
        ,asset.purchase_date
        ,asset.purchase_order
        ,asset.value
        ,asset.type
        ,asset.notes
        ,laptop.manufacturer
        ,laptop.model
        ,laptop.serial_number
        ,laptop.esc
        ,laptop.user
        ,laptop.prev_user
        ,laptop.warranty
        FROM asset
        LEFT JOIN laptop
        ON laptop.id = asset.id
        WHERE asset.id = 93120
        ");
        break;  
    case "Desktop":
        //do some stuff that involves a mysql query
        mysql_query("
        SELECT asset.id
        ,asset.company
        ,asset.location
        ,asset.purchase_date
        ,asset.purchase_order
        ,asset.value
        ,asset.type
        ,asset.notes
        ,desktop.manufacturer
        ,desktop.model
        ,desktop.serial_number
        ,desktop.esc
        ,desktop.user
        ,desktop.prev_user
        ,desktop.warranty
        FROM asset
        LEFT JOIN desktop
        ON desktop.id = asset.id
        WHERE asset.id = 93120
        ");
        break;  
}

?>

到目前为止，我已经能够将asset.type转换为$ type。 我将如何获取其余的变量（laptop.model到$ model，asset.notes到$ notes等）？

谢谢。

Answer 1

$sql = "YOUR QUERY";
$res = mysql_query($sql);

while($row = mysql_fetch_assoc($res))
{
    echo 'Start Record<br />';
    echo $row['type'].'<br />';
    echo $row['company'].'<br />';
    echo $row['location'].'<br />';
    echo 'End Record<br /> <br />';
}

尝试一下，看看会得到什么，然后您可以根据需要使用数据；

您可能还需要查看mysql_fetch_array ， mysql_fetch_row或mysql_fetch_object 。 选择最适合您的需求。

Answer 2

您正在执行的操作无效：

$type = mysql_query("
SELECT asset.type
From asset
WHERE asset.id = 93120
")

这会将结果集放入$ type，而不是类型本身。

完成之后，您需要获取一行，然后获取字段：

$result = mysql_query("
SELECT asset.type
From asset
WHERE asset.id = 93120
")
if($row = mysql_fetch_object($result)){
  $type = $row->type;

  // NOW you have the type in $type. Do something similar with the rest of the queries.
}

Answer 3

我猜你会想做这样的事情：

$result = mysql_query($query);
$i = array();
while ($data = mysql_fetch_assoc($result)) {
 $i[] = $data;
}

这将使$ ia多维数组包含所有查询数据，您可以通过执行以下操作来使用它

foreach ($i as $key => $value) {
  echo $value['model'];
  echo $value['serial'];
  etc......
}

Answer 4

请参见：

http://us2.php.net/manual/en/mysql.examples-basic.php

http://us2.php.net/manual/en/function.mysql-query.php

Answer 5

您可以看看http://php.net/manual/en/function.mysql-query.php 。

mysql_query返回包含查询结果的资源。 可以通过以下方式读取此资源：

$sql_result = mysql_query( [here your stuff] );
while ($row = mysql_fetch_assoc($sql_result)){
  echo $row['model']; // prints the model
}

因此，您必须遍历结果并遍历每一row 。 那么row是一个包含单个字段的关联数组。

希望这可以帮助。

Answer 6

它只能是一台台式computers

所以你的代码就是：

<?php
//make database connect
mysql_connect("localhost", "asset_db", "asset_db") or die(mysql_error());
mysql_select_db("asset_db") or die(mysql_error());

$sql = "SELECT a.id, a.company, a.location, a.purchase_date, a.purchase_order,
               a.value, a.type, a.notes, c.manufacturer, c.model, 
               c.serial_number, c.esc, c.user, c.prev_user, c.warranty
        FROM asset a
        LEFT JOIN computers c
        ON c.id = a.id
        WHERE a.id = 93120";
$res = mysql_query($sql) or trigger_error(mysql_error().$sql);
$data=array();
while($row = mysql_fetch_assoc($res)) $data[] = $row;
// now $data array contains all the rows returned
?>

每次您看到代码加倍时，就会知道您做错了什么。