我有一个工作很好的链选择,但我希望能够将数据从搜索记录表单推送到包含链选择的表单,如何在此代码中使用SELECTED = SELECTED?
我的select.class.php
<?php
class SelectList {
public function ShowMake($searchboilermake) {
include "./core/cnn.php";
$sql = "SELECT * FROM blr_boilermake";
$res = mysql_query($sql);
$boilermake = '<option value=""></option>';
while($row = mysql_fetch_array($res)) {
$boilermake .= '<option value="' . $row['BoilerID'] . '"';
$boilermake .= ($row['BoilerID']==$searchboilermake) ? " selected='selected' " : "";
$boilermake .= '>' . $row['BoilerMake'] . '</option>';
}
return $boilermake;
}
public function ShowModel() {
include "../core/cnn.php";
$sql = "SELECT * FROM blr_boilermodel WHERE BoilerModel IS NOT NULL AND BoilerID = $_POST[id]";
$res = mysql_query($sql);
$boilermodel = '<option value="0"></option>';
while($row = mysql_fetch_array($res)) {
$boilermodel .= '<option value="' . $row['ModelID'] . '">' . $row['BoilerModel'] . '</option>';
}
return $boilermodel;
}
}
$opt = new SelectList();
?>从我页面上的表单中,我使用它来拉入boilermake字段的select.class.php
<label>Manufacturer</label>
<select class="select2_category form-control" data-placeholder="Choose a Category" tabindex="1" id="boilermanufacturer" name="boilermanufacturer">
<?php echo $opt->ShowMake($_REQUEST['boilermanufacturer']); ?>
</select>这是我的boilermodel字段
<label>Model</label>
<select class="select2_category form-control" data-placeholder="Choose a Category" tabindex="1" id="boilermodel" name="boilermodel">
<option value=""></option>
</select>在其他更直接和更标准的选择中,我在下面使用了这个,但是我如何才能使下面的选择适应我的链选择?
<div class="form-group">
<label>Fuel Type</label>
<?php $fueltype = db::getInstance()->query('SELECT * FROM lkup_fueltype');
if(!$fueltype->count()) {
echo 'Problem';
} else { ?>
<select class="select2_category form-control" data-placeholder="Choose a Category" tabindex="1" id="propertyfueltype" name="propertyfueltype">
<?php foreach ($fueltype->results() as $fueltype) { ?>
<option value="<?php echo $fueltype->ID; ?>"<?php echo $fueltype->ID == $searchboilerfueltype ? "selected" : ""; ?>><?php echo $fueltype->PropertyFuelType; ?></option> <?php } } ?>
</select>
我尝试使用它,但当我运行代码时,它没有选择选定的行
$boilermake .= '<option value="' . $row['BoilerID'] . '" "' . $row['BoilerID'] . '" == $searchboilermake ? "selected" : "">' . $row['BoilerMake'] . '</option>';我应该做些什么/或者我能做些什么才能让它正常工作?
发布于 2014-05-20 07:28:42
问题是您试图在字符串中使用布尔/三元表达式。这显然不会在这样的字符串中进行计算。
$boilermake .= '<option value="' . $row['BoilerID'] . '"';
$boilermake .= ($row['BoilerID']==$searchboilermake) ? " selected='selected' " : "";
$boilermake .= '>' . $row['BoilerMake'] . '</option>';编辑:您需要修改函数ShowMake以将$searchboilermake作为参数传入。
函数定义:
public function ShowMake($searchboilermake) {对函数的调用:
<?php echo $opt->ShowMake($_REQUEST['boilermanufacturer']); ?>https://stackoverflow.com/questions/23748012
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