獲得兩個腳本共同努力

Question

我想，這樣的第一個結果確定第二個的輸出使用下面的腳本。

<? 
$db = mysql_connect('localhost','username','pass') or die("Database error"); 
mysql_select_db('dbname', $db); 

$query = "SELECT pool FROM winners"; 
$result = mysql_query($query) or die(mysql_error()); 

while ($row = mysql_fetch_array($result)) 

if ($row['pool'] % 2) 
{ 
echo "<h4>Result 1</h4>"; 
echo "<br />"; 
} 
else 
{ 
echo "<h4>Result 2</h4>"; 
echo "<br />"; 
} 
?> 


<?php 

$db = mysql_connect('localhost','username','pass') or die("Database error"); 
mysql_select_db('dbnamesameasother', $db); 


$query2 = "SELECT * FROM comments"; 
$result2 = mysql_query($query2); 

while ($row2 = mysql_fetch_assoc($result2)) 

if ($row2['commentid'] % 2 ==0) 
{ 
echo $row2['name']; 
echo "<br />"; 
} 
else 
{ 
echo $row2['name']; 
} 
?> 

如果第一個腳本選擇結果1所以基本上我只想回音聯想與結果的名稱。這些名字通過紀念相關聯，如果奇怪的紀念是結果2，甚至紀念是結果1.有沒有辦法做到這一點，而不使用工會聲明？

Answer 1

它建議你創建一個函數，然後調用it.something這樣的：

<?php 
$db = mysql_connect('localhost','username','pass') or die("Database error"); 
mysql_select_db('dbname', $db); 

$query = "SELECT pool FROM winners"; 
$result = mysql_query($query) or die(mysql_error()); 

while ($row = mysql_fetch_array($result)) 

if ($row['pool'] % 2) 
{ 
echo "<h4>Result 1</h4>"; 
$names = get_names(1); 
    foreach ($names as $name) { 
    echo $name . "<br/>"; 
    } 
} 
else 
{ 
echo "<h4>Result 2</h4>"; 
$names = get_names(0); 
    foreach ($names as $name) { 
    echo $name . "<br/>"; 
    } 
} 

Function get_names($pool_result) 
{ 
    $name_array = array(); 
    $db = mysql_connect('localhost','username','pass') or die("Database error"); 
    mysql_select_db('dbnamesameasother', $db); 

    $query = "SELECT * FROM comments WHERE commentid % 2 = $pool_result"; 
    $result = mysql_query($query); 

    while ($row = mysql_fetch_array($result)) 
    { 
    array_push($name_array , $row['name']); 
    } 
    return $name_array; 
} 
?> 

}