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[英]Combine these scripts to work together, Javascript/AJAX and PHP
[英]Getting two scripts to work together
我正在尝试使用以下脚本,以便第一个脚本的结果确定第二个脚本的输出。
<?
$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'];
}
?>
因此,基本上,如果第一个脚本选择结果1,我只想回显与此结果关联的名称。 这些名称由commentid关联,其中奇数commentid将是结果2,偶数commentid将是结果1。是否有任何方法可以在不使用union语句的情况下做到这一点?
建议您先创建一个函数,然后调用它。类似这样:
<?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;
}
?>
}
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