【发布时间】:2016-04-10 11:30:12
【问题描述】:
所以我有这个测试代码到一个主要项目。这段代码完美地连接到数据库,几乎完成了我设定的一切。但是,当我从下拉列表中选择最后一个值时,它不会无缘无故地给我它的价值。 这是我想要的示例。
这里是一个“错误”的例子。
这是完整的代码(连接除外)。
<?php
include_once('ligacao.php');
//Check connection
if (!$link) {
die("Connection failed: " . mysqli_connect_error());
}
if (!empty($_GET['id'])) {
$pesq = $_GET['id'];
$atual = $_GET['atual'];
//Distrito e concelho
$sql = ("SELECT distrito,concelho FROM concelhos WHERE codg_cc = '$pesq'");
$result = mysql_query($sql, $link);
while ($row = mysql_fetch_assoc($result)) {
$dist = $row['distrito'];
$conc = $row['concelho'];
}
//echo ($atual);
if (isset($atual) and strlen($atual) > 0) {
$sql3 = ("SELECT DISTINCT new_name_freg,cod_freg_new FROM freguesias WHERE cod_freg_new = '$atual'");
$result3 = mysql_query($sql3, $link);
$sql2 = ("SELECT old_name_freg,cod_freg_old FROM freguesias WHERE cod_freg_new = '$atual'");
$result2 = mysql_query($sql2, $link);
} else {
$sql3 = ("SELECT DISTINCT new_name_freg,cod_freg_new FROM freguesias WHERE codg_cc = '$pesq' ORDER BY new_name_freg");
$result3 = mysql_query($sql3, $link);
}
mysql_free_result($result);
}
/* $atual=$_GET['atual']; // This line is added to take care if your global variable is off
if(isset($atual) and strlen($atual) > 0){
//echo ('hello');
$sql2=("SELECT old_name_freg,cod_freg_old FROM freguesias WHERE cod_freg_new = '$atual'");
$result2 = mysql_query($sql2, $link);
} */
/*
} // else{$quer="SELECT DISTINCT subcategory,subcat_id FROM subcategory order by subcategory"; }
////////// end of query for second subcategory drop down list box ///////////////////////////
*/
mysqli_close($ligar);
?>
<html>
<head>
<script type="text/javascript">
/*function reload(form)
{
val=$pesq;
self.location='Formulario2.php?id=' + val ;
} */
function reload3(form)
{
var concelho = '<?= $pesq ?>'; //passa a variavel de php para o javascript
//alert(concelho);
//alert(element.defaultValue);
//alert(self.location);
val2 = (form.value);
self.location = 'Formulario2.php?id=' + concelho + '&atual=' + val2;
}
</script>
</head>
<body onload=enable_text(false);>
<div >
<table>
<form name=f1 action="" >
<td>
<tr>
<td>ID:</td> <td><input type="number" name="id" value="<?php echo $pesq; ?>"></td>
<td><input type="submit" value="Obter informações"></td>
</tr>
</td>
</form>
</table>
<form name=f2 action="" >
<table>
<tr>
<td>Distrito:</td><td ><input type="text" name="distrito" maxlength="25" value="<?php echo $dist; ?>"></td>
<td>Concelho:</td> <td><input type="text" name="concelho" maxlength="25" value="<?php echo $conc; ?>"></td><br>
</tr>
<tr>
<td>Freguesias:</td>
<td>Designação atual :</td> <td><select name="atual" onchange="reload3(this)"><option value=''>Selecione uma freguesia</option>
<?php
while ($row3 = mysql_fetch_assoc($result3)) {
echo "<option size=30 selected value=" . $row3["cod_freg_new"] . ">" . $row3["new_name_freg"] . "</option>";
}
mysql_free_result($result3);
?>
</select></td>
<td>Designação antiga :</td> <td><select name="antiga" onchange="reload3(this"><option value=''>Select One</option>
<?php
while ($row2 = mysql_fetch_assoc($result2)) {
echo "<option size=30 value=" . $row2["cod_freg_old"] . ">" . $row2["old_name_freg"] . "</option>";
}
mysql_free_result($result2);
?>
</select></td>
</form>
</tr>
<tr>
<td>Número da mesa:</td> <td><input type="number" name="mesa" min="1" max="15"></td><br>
<td>Localização da mesa:</td> <td><input type="text" name="local" maxlength="40"></td><br>
</tr>
<td>Todos os eleitores desta frequesia: <input type="checkbox" name=todos onclick="enable_text(this.checked)" ></td>
</tr>
<tr>
<td>Nº de eleitor inicial do caderno </td><td><input type=text name=intini1 maxlength=5></td><td><input type=number name=intini2 maxlength=5></td>
<td>Nº de eleitor final do caderno </td><td><input type=text name=intfim1 maxlength=5></td><td><input type=number name=intfim2 maxlength=5></td>
</tr>
<tr><br>
<td>Nota:</td> <td><input type="text" name="nota" maxlength="40"></td><br>
</tr>
</table>
</form>
</div>
</body>
</html>
如果您对代码有任何疑问,或者需要我澄清我的问题,请不要害羞地告诉我。
P.S:我知道我应该使用 mysqli 而不是 myslq_ 并且还准备好声明,但我试图专注于真正的问题而不是安全性。如果我的英语不好,请原谅我。
这里是请求的html源代码。
<form name="f2" action="">
<br>
<br>
<br>
<br>
<br>
<table>
<tbody>
<tr>
<td>Distrito:</td>
<td><input type="text" name="distrito" maxlength="25" value="Faro"></td>
<td>Concelho:</td>
<td><input type="text" name="concelho" maxlength="25" value="Albufeira"></td>
</tr>
<tr>
<td>Freguesias:</td>
<td>Designação atual :</td>
<td><select name="atual" onchange="reload3(this)">
<option value="">Selecione uma freguesia</option>
<option size="30" selected="" value="080106">Albufeira e Olhos de Ãgua</option>
<option size="30" selected="" value="080104">Ferreiras</option>
<option size="30" selected="" value="080102">Guia</option>
<option size="30" selected="" value="080103">Paderne</option>
</select></td>
<td>Designação antiga :</td>
<td><select name="antiga" onchange="reload3(this">
<option value="">Select One</option>
</select></td>
</tr>
<tr>
<td>Número da mesa:</td>
<td><input type="number" name="mesa" min="1" max="15"></td>
<td>Localização da mesa:</td>
<td><input type="text" name="local" maxlength="40"></td>
</tr>
<tr>
<td>Todos os eleitores desta frequesia:
<input type="checkbox" name="todos" onclick="enable_text(this.checked)"></td>
</tr>
<tr>
<td>Nº de eleitor inicial do caderno </td>
<td><input type="text" name="intini1" maxlength="5"></td>
<td><input type="number" name="intini2" maxlength="5"></td>
<td>Nº de eleitor final do caderno </td>
<td><input type="text" name="intfim1" maxlength="5"></td>
<td><input type="number" name="intfim2" maxlength="5"></td>
</tr>
<tr>
<td>Nota:</td>
<td><input type="text" name="nota" maxlength="40"></td>
</tr>
</tbody>
</table>
</form>
这是所要求的 javascript 代码。
function reload3(form)
{
var concelho = '<?= $pesq ?>'; //passa a variavel de php para o javascript
//alert(concelho);
//alert(element.defaultValue);
//alert(self.location);
val2=(form.value);
self.location='Formulario2.php?id=' + concelho + '&atual=' + val2;
}
更新: 我工作的人使用 JQuery 找到了一个更好的解决方案,从而解决了这个问题。感谢所有帮助过的人。
【问题讨论】:
-
请显示生成的html,可能该值格式错误,例如其中包含引号或其他内容
-
尝试在选项中的值周围添加引号。
-
图片有助于解释问题,但请您按要求添加 html 源代码 - 例如右键单击 > 在浏览器中查看源代码并粘贴表单 html
-
@jcubic 谢谢你的帮助。我已经添加了引号,但这不是解决方案。
-
@Steve 谢谢你的帮助。我在答案中添加了源代码。很抱歉没听懂你说的。
标签: javascript php html mysql