因此,我仅使用PHP和Ajax制作饼图。但是我看不到必须如何进一步编码才能使其正常工作(请参见第一张图片)。因此,当某人单击按钮ABC或D时,必须看到(不加载页面)您已对其中之一投了赞成票,并且还必须在图表中看到。就是这样!第二张图片显示了我的数据库。
在我忘记告诉我数据库中没有要更改的图片之前。
我希望你们中的一些可以帮助我。我的一些代码:
<p><h1>Breng jou stem uit</h1></p><br />
<form action = "<?php
echo $_SERVER['PHP_SELF'];
?>" method = "GET">
<button type="button" name="a">Partij A</button><br />
<button type="button" name="b">Partij B</button><br />
<button type="button" name="c">Partij C</button><br />
<button type="button" name="d">Partij D</button>
</form>
<?php
// Connects to your Database
include('../../../connection.php');
$sql = mysql_query("SELECT * FROM votes");
while ($row = mysql_fetch_array($sql)) {
echo $partijA = $row['partijA'];
$partijB = $row['partijB'];
$partijC = $row['partijC'];
$partijD = $row['partijD'];
if (isset($_GET['a'])) {
echo $resultA = $partijA + 1;
} else {
echo "1";
}
$resultB = $partijB + 1;
$resultC = $partijC + 1;
$resultD = $partijD + 1;
}
// Name of our cookie
$cookie = "Voted";
// A function to display our results - this refrences vote_pie.php which we will also make
function pie()
{
$data = mysql_query("SELECT * FROM votes") or die(mysql_error());
$result = mysql_fetch_array($data);
$total = $result[partijA] + $result[partijB] + $result[partijC] + $result[partijD];
$one = round(360 * $result[partijA] / $total);
$two = round(360 * $result[partijB] / $total);
$per1 = round($result[partijA] / $total * 100);
$per2 = round($result[partijB] / $total * 100);
$per3 = round($result[partijC] / $total * 100);
$per4 = round($result[partijD] / $total * 100);
echo "<img src=vote_pie.php?one=" . $one . "&two=" . $two . "><br/>";
Echo "<font color=000000>Partij A</font> = $result[partijA] votes = $per1 <br />
<font color=000000>Partij B</font> = $result[partijB] votes = $per2 <br />
<font color=000000>Partij C</font> = $result[partijC] votes = $per3 <br />
<font color=000000>Partij D</font> = $result[partijD] votes = $per4 <br />";
}
// displays the poll results
pie();
?>
您的代码中存在许多问题,但导致您遇到实际问题的一个问题是,您的函数pie
不会递增从数据库中获取的值,而是在函数外部完成,而是递增的值在那里不使用:
function pie()
{
//this query was already performed, pass the result resource to this function, don't re-run the query
$data = mysql_query("SELECT * FROM votes") or die(mysql_error()); //google or die must die
$result = mysql_fetch_array($data);//mysql is deprecated
//array keys need to be quoted
$total = $result[partijA] + $result[partijB] + $result[partijC] + $result[partijD];
$one = round(360 * $result[partijA] / $total);
$two = round(360 * $result[partijB] / $total);
$per1 = round($result[partijA] / $total * 100);
$per2 = round($result[partijB] / $total * 100);
$per3 = round($result[partijC] / $total * 100);
$per4 = round($result[partijD] / $total * 100);
//functions return, they don't echo
echo "<img src=vote_pie.php?one=" . $one . "&two=" . $two . "><br/>";
Echo "<font color=000000>Partij A</font> = $result[partijA] votes = $per1 <br />
<font color=000000>Partij B</font> = $result[partijB] votes = $per2 <br />
<font color=000000>Partij C</font> = $result[partijC] votes = $per3 <br />
<font color=000000>Partij D</font> = $result[partijD] votes = $per4 <br />";
}
我随意添加一些注释来指出代码中的几个问题。
只需删除在pie
函数外部查询和递增的代码,然后让函数完成所有工作即可。向其传递所需的值(即使$_GET
它返回结果,而不是回显它)。
我将重写函数以接受这样的参数(此时不使用mysql):
function getPie(array $params, $connection)
{
//added limit 1, seeing as you're only processing the first result
$result = mysql_query($connection, 'SELECT * from votes LIMIT 1');
$row = mysql_fetch_assoc($result);
if (isset($params['a'])) {
$row['partijA'] += 1;
} else if (isset($params['b'])) {
$row['partijB'] += 1;
} //add more else's for all parties
$percentages = [];//array
foreach ($row as $key => $val) {
$percentages[$key] = round(($val/$total) * 100);
}
$total = array_sum($row);//total of all votes
$markup = '<img src=vote_pie.php?one=' .
$percentages['partijA']*360 . '&two=' . $percentages['partijB']*360 . '><br>';
foreach ($percentages as $k => $perc) {
$markup .= '<font color=000000>Partij ' . str_replace('partij', $k) . '</font> = '.
$row[$key] . ' votes = ' . $perc . '<br>';
}
return $markup;
}
然后像下面这样调用此函数:
if ($_GET) {//if a get form was submitted
echo getPie($_GET, $db);//where $db holds the mysql resource
}
不过,这只是一个快速解决方案...还有很长的路要走...
快速查看我所做的实际更改:
$_GET
在这种情况下为值)和要使用的db连接。round($resutl['key']/total*100)
我使用了循环,而不是重复,所以我只需要编写一次该语句,并将其应用于所有值$row
or$percentages
数组进行迭代(它们都具有相同的键) ,并填写具体值。这也意味着我可以安全地添加一个聚会,而不必更改创建标记的代码本文收集自互联网,转载请注明来源。
如有侵权,请联系 [email protected] 删除。
我来说两句