【发布时间】:2017-02-10 07:00:38
【问题描述】:
这就是我需要的:当我选择“10 比索”之类的选项时...
Image One
我需要更改其他几个选择选项上的 optgroup 标签,例如: 我想要更改的信息为“Recheios Especiais:R$10”,例如更改为“Recheios Especiais:R$20”。当我选择“30 比索”时,将标签更改为“Recheios Especiais:R$40”。
这是我的代码:(我使用的是 select2 插件,但我认为这与问题无关)
猜猜这个解决方案与这个非常相似,但我不需要更改<div> 的内容,而是需要更改<optgroup label=""> 的内容:(评论上有链接)
<form class="form" method='post' id='formBoloMesa' action="custom_includes/orderBolo.php">
<div class="form-group">
<span class="resume-subtitle">Escolha o tamanho</span>
<select name="tamanho" class="form-control select2-list select-options-1" data-placeholder="Clique aqui para escolher" multiple>
<optgroup label="Tamanho">
<?php
include ('custom_includes/connect.php');
$strTamanho = "SELECT * FROM bolo_mesa_itens WHERE tipo='tamanho'";
$rsTamanho = mysql_query($strTamanho);
while($rowTamanho = mysql_fetch_array($rsTamanho)) {
echo '<option value="' . $rowTamanho['nome'] . '">' . $rowTamanho['nome'] . '</option>';
}
?>
</optgroup>
</select>
</div>
<div class="form-group">
<span class="resume-subtitle">
Escolha o modelo
</span>
<select name="modelo" class="form-control select2-list select-options-1" data-placeholder="Clique aqui para escolher" multiple>
<optgroup label="Tipo de Bolo">
<?php
$strModelo = "SELECT * FROM bolo_mesa_itens WHERE tipo='modelo'";
$rsModelo = mysql_query($strModelo);
while($rowModelo = mysql_fetch_array($rsModelo)) {
echo '<option value="' . $rowModelo['nome'] . '">' . $rowModelo['nome'] . '</option>';
}
?>
</select>
</div>
<div class="form-group">
<span class="resume-subtitle">Escolha a Massa</span>
<select name="massa" class="form-control select2-list select-options-1" data-placeholder="Clique aqui para escolher" multiple>
<optgroup label="Massas">
<?php
$strMassas = "SELECT * FROM bolo_mesa_itens WHERE tipo='massa'";
$rsMassas = mysql_query($strMassas);
while($rowMassas = mysql_fetch_array($rsMassas)) {
echo '<option value="' . $rowMassas['nome'] . '">' . $rowMassas['nome'] . '</option>';
}
?>
</optgroup>
</select>
</div>
<div class="form-group">
<span class="resume-subtitle">Escolha 3 recheios</span>
<select name="recheio1" class="form-control select2-list select-options-1" data-placeholder="Escolha o 1º Recheio" multiple>
<optgroup label="Recheios Comuns">
<?php
$strRecheios = "SELECT * FROM bolo_mesa_itens WHERE tipo='recheio'";
$rsRecheios = mysql_query($strRecheios);
while($rowRecheios = mysql_fetch_array($rsRecheios)) {
echo '<option value="' . $rowRecheios['nome'] . '">' . $rowRecheios['nome'] . '</option>';
}
?>
</optgroup>
<optgroup label="Recheios Especiais (+R$10)">
<?php
$strRecheiosPremium = "SELECT * FROM bolo_mesa_itens WHERE tipo='recheio' AND premium='1'";
$rsRecheiosPremium = mysql_query($strRecheiosPremium);
while($rowRecheiosPremium = mysql_fetch_array($rsRecheiosPremium)) {
echo '<option value="' . $rowRecheiosPremium['nome'] . '">' . $rowRecheiosPremium['nome'] . '</option>';
}
?>
</optgroup>
</select>
【问题讨论】:
-
猜猜这个解决方案与这个非常相似,但我不需要更改的内容,而是需要更改
标签: javascript select2 jquery-selectbox