使用复选框问题显示数据库中的值
使用复选框问题显示数据库中的值
提问于 2013-09-12 22:29:12
关于使用复选框显示数据库中的数据，我有一个问题。我可以展示它们，但它们不止一次显示。我只需要展示一次。如果有更多的结果相同的东西，那么我应该只显示其中一个。如果我选择了多个复选框，就会得到这样的结果
ISO 27005:2008
NIST SP 800 - 30
OCTAVE
CobIT
CRAMM
FMEA
FRAP
EBIOS
RuSecure
ALE
Cobra
MEHARI
ISO 27005:2008
NIST SP 800 - 30
OCTAVE
CobIT
CRAMM
FMEA
FRAP
EBIOS
RuSecure
ALE
Cobra
MEHARI
ISO 27005:2008
NIST SP 800 - 30
OCTAVE
CobIT
CRAMM
FMEA
FRAP
EBIOS
RuSecure
ALE
Cobra
MEHARI我只需要一个结果，而不是同样的结果。我需要某种过滤器或什么东西来比较所有列出的，并只选择一个实例。每个显示结果都是选中的每个复选框的结果。如果我选择其中的两个，那么将显示每个结果的两个结果，我只想对每个结果进行一个合并，如下所示：
ISO 27005:2008
NIST SP 800 - 30
OCTAVE
CobIT
CRAMM
FMEA
FRAP
EBIOS
RuSecure
ALE
Cobra
MEHARI 或者其他符合表中条件的组合，对于删除重复的值有什么帮助或建议吗？，我是一个初学者，对我的语言感到非常抱歉。
This is what I've got.lol.php This is where all the action is. I use if statements to check for checked checkboxes.
<?php
if (isset($_POST['submit'])) {
if (isset($_POST['chkbox1']) || (isset($_POST['chkbox2'])) || (isset($_POST['chkbox3']))) {
if ((isset($_POST['chkbox1']) && (isset($_POST['chkbox2'])))) {
$upit = mysql_query("SELECT ime FROM kriteriji where identifikacija = 1 and analiza = 1");
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
echo $cijena . "<br>";
}
}
if ((isset($_POST['chkbox1'])) && (isset($_POST['chkbox3']))) {
$upit = mysql_query("SELECT ime FROM kriteriji where identifikacija = 1 and evaluacija = 1");
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
echo $cijena . "<br>";
}
}
if ((isset($_POST['chkbox2'])) && (isset($_POST['chkbox3']))) {
$upit = mysql_query("SELECT ime FROM kriteriji where analiza = 1 and evaluacija = 1");
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
echo $cijena . "<br>";
}
}
if (isset($_POST['chkbox1'])) {
$upit = mysql_query("SELECT ime FROM kriteriji where identifikacija = 1");
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
echo $cijena . "<br>";
}
}
if (isset($_POST['chkbox2'])) {
$upit = mysql_query("SELECT ime FROM kriteriji where analiza = 1");
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
echo $cijena . "<br>";
}
}
if (isset($_POST['chkbox3'])) {
$upit = mysql_query("SELECT ime FROM kriteriji where evaluacija = 1");
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
echo $cijena . "<br>";
}
}
} else {
$upit = mysql_query("SELECT ime FROM kriteriji where identifikacija = 1 and analiza = 1 and evaluacija = 1");
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
echo $cijena . "<br>";
}
}
}
?>
html file
<!doctype html>
<html>
<head>
<?php
//spajanje na server
$spajanje = mysql_connect("localhost", "root", "") or die("Nije se mouće povezati na server :("); //spajanje na server
//spajanje na bazu
$baza = mysql_select_db("magistarski") or die("Nije se moguće povezati na bazu podataka :(");
?>
<meta charset="utf-8">
<title>Magistarski</title>
</head>
<body>
<form action="lol.php" method="post">
<p>
<table width="766" height="418" border="0">
<caption>
<h1>Choose...</h1>
</caption>
<tr>
<td width="203" height="203"><h3 align="left"><em>Risk Assessment</em></h3>
<div align="justify"><em>
<input type="checkbox" name="chkbox1" value="risk_id" />
Risk Identification<br>
<input type="checkbox" name="chkbox2" value="risk_an" />Risk Analysis<br>
<input type="checkbox" name="chkbox3" value="risk_eval" />Risk Evaluation<br>
</em></div></td>
<td width="194"><h3 align="left"><em>Risk management</em></h3>
<div align="justify"><em>
<input type="checkbox" name="chkbox4" value="risk_ass" />
Risk Assessment<br>
<input type="checkbox" name="chkbox5" value="risk_acc" />Risk Acceptance<br>
<input type="checkbox" name="chkbox6" value="risk_com" />Risk Communication<br>
</em></div></td>
<td width="212"><h3 align="justify"><em>Price</em></h3>
<div align="justify"><em>
<input type="checkbox" name="chkbox7" value="free" />
Free<br>
<input type="checkbox" name="chkbox8" value="notfree" />Not Free<br>
</em></div></td>
</tr>
<tr>
<td height="181"><h3 align="justify"><em>Organization</em></h3>
<div align="justify"><em>
<input type="checkbox" name="chkbox9" value="sme" />SME<br>
<input type="checkbox" name="chkbox10" value="large" />Large<br>
<input type="checkbox" name="chkbox11" value="goverment" />Goverment<br>
<input type="checkbox" name="chkbox12" value="profit" />Profit<br>
<input type="checkbox" name="chkbox13" value="nonprofit" />Non - profit<br>
</em></div></td>
<td><h3 align="justify"><em>Skills</em></h3>
<div align="justify"><em>
<input type="checkbox" name="chkbox14" value="basic" />Basic<br>
<input type="checkbox" name="chkbox15" value="standard" />Standard<br>
<input type="checkbox" name="chkbox16" value="specialist" />Specialist<br>
</em></div></td>
<td><h3 align="justify"><em>Documented</em></h3>
<div align="justify"><em>
<input type="checkbox" name="chkbox17" value="low" />
Low<br>
<input type="checkbox" name="chkbox18" value="good" />Good<br>
<input type="checkbox" name="chkbox19" value="high" />High<br>
</em></div></td>
</tr>
</table>
</p>
<div>
<input type="submit" name="submit" value="Show" />
</div>
</form>
</body>
</html>回答 1
Stack Overflow用户
发布于 2013-09-12 23:20:23
如果我把你说对了，你就用' if‘表达式检查发送的数据，这样做是错误的。我认为您应该首先获得所有必需的字段，然后编写查询。试着做这样的事情：
if (isset($_POST['gumbposlan'])) {
$whereString = '';
if (isset($_POST['chkbox'])){
foreach ($_POST['chkbox'] as $value) {
if (empty($whereString))
$whereString .= "where `$value` = 1 ";
else
$whereString .= "and where `$value` = 1 ";
}
}
$query = "SELECT DISTINCT `ime` FROM `kriteriji` $whereString";
$upit = mysql_query($query);
while ($red = mysql_fetch_array($upit)) {
$cijena = $red['ime'];
if ($cijena < 1)
echo $cijena . "<br>";
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/18775324
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