php 使用php mysql错误从数据库中选择的值进入下拉选择框选项
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selected value get from db into dropdown select box option using php mysql error
提问by Php Gemini
I need to get selected value from db into select box. please, tell me how to do it. Here is the code. Note: 'options' value depends on the category.
我需要从 db 中获取选定的值到选择框中。请告诉我怎么做。这是代码。注意：'options' 值取决于类别。
<?php
$sql = "select * from mine where username = '$user' ";
$res = mysql_query($sql);
while($list = mysql_fetch_assoc($res)){
$category = $list['category'];
$username = $list['username'];
$options = $list['options'];
?>
<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
<option value="0">Please Select Option</option>
<option value="PHP">PHP</option>
<option value="ASP">ASP</option>
</select>
<?php
}
?>
回答by rakeshjain
I think you are looking for below code changes:
我认为您正在寻找以下代码更改：
<select name="course">
<option value="0">Please Select Option</option>
<option value="PHP" <?php if($options=="PHP") echo 'selected="selected"'; ?> >PHP</option>
<option value="ASP" <?php if($options=="ASP") echo 'selected="selected"'; ?> >ASP</option>
</select>
回答by independent.guru
The easiest way I can think of is the following:
我能想到的最简单的方法如下：
PHP
PHP
<?php
$selection=array('PHP','ASP');
echo '<select>
<option value="0">Please Select Option</option>';
foreach($selection as $selection){
$selected=($options == $selection)? "selected" : "";
echo '<option '.$selected.' value="'.$selection.'">'.$selection.'</option>';
}
echo '</select>';
The code basically places all of your options in an array which are called upon in the foreach loop. The loop checks to see if your $options variable matches the current selection it's on, if it's a match then $selected will = selected, if not then it is set as blank. Finally the option tag is returned containing the selection from the array and if that particular selection is equal to your $options variable, it's set as the selected option.
代码基本上将所有选项放在一个数组中，这些选项在 foreach 循环中被调用。循环检查您的 $options 变量是否与它所在的当前选择匹配，如果匹配，则 $selected 将 = 选择，如果不匹配，则将其设置为空白。最后返回包含数组中选择的选项标签，如果该特定选择等于您的 $options 变量，则将其设置为所选选项。
回答by chirag ode
for example ..and please use mysqli() next time because mysql() is deprecated.
例如..下次请使用 mysqli() 因为 mysql() 已被弃用。
<?php
$select="select * from tbl_assign where id='".$_GET['uid']."'";
$q=mysql_query($select) or die($select);
$row=mysql_fetch_array($q);
?>
<select name="sclient" id="sclient" class="reginput"/>
<option value="">Select Client</option>
<?php $s="select * from tbl_new_user where type='client'";
$q=mysql_query($s) or die($s);
while($rw=mysql_fetch_array($q))
{ ?>
<option value="<?php echo $rw['login_name']; ?>"<?php if($row['clientname']==$rw['login_name']) echo 'selected="selected"'; ?>><?php echo $rw['login_name']; ?></option>
<?php } ?>
</select>
回答by Siddharth Shukla
Select value from drop down.
从下拉列表中选择值。
<select class="form-control" name="category" id="sel1">
<?php
foreach($data as $key =>$value){
?>
<option value="<?php echo $data[$key]->name; ?>"<?php if($id_name[0]->p_name==$data[$key]->name) echo 'selected="selected"'; ?>><?php echo $data[$key]->name; ?></option>
<?php } ?>
</select>
回答by razvan
BEST code and simple
最好的代码和简单
<select id="example-getting-started" multiple="multiple" name="category">
<?php
$query = "select * from mine";
$results = mysql_query($query);
while ($rows = mysql_fetch_assoc(@$results)){
?>
<option value="<?php echo $rows['category'];?>"><?php echo $rows['category'];?></option>
<?php
}
?>
</select>
回答by K. Raza
THE EASIEST SOLUTION
最简单的解决方案
It will add an extra in your options but your problem will be solved.
它会在您的选项中添加额外的内容，但您的问题将得到解决。
<?php
if ($editing == Yes) {
echo "<option value=\".$MyValue.\" SELECTED>".$MyValue."</option>";
}
?>
回答by narinder kumar
You can also do like this ....
你也可以这样做......
<?php $countryname = $all_meta_for_user['country']; ?>
<select id="mycountry" name="country" class="user">
<?php $myrows = $wpdb->get_results( "SELECT * FROM wp_countries order by country_name" );
foreach($myrows as $rows){
if( $countryname == $rows->id ){
echo "<option selected = 'selected' value='".$rows->id."'>".$rows->country_name."</option>";
} else{
echo "<option value='".$rows->id."'>".$rows->country_name."</option>";
}
}
?>
</select>
回答by 7ani9
USING POD
使用 POD
<?php
$username = "root";
$password = "";
$db = "db_name";
$dns = "mysql:host=localhost;dbname=$db;charset=utf8mb4";
$conn = new PDO($dns,$username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "select * from mine where username = ? ";
$stmt1 = $conn->prepare($sql);
$stmt1 -> execute(array($_POST['user']));
$all = $stmt1->fetchAll(); ?>
<div class="controls">
<select data-rel="chosen" name="degree_id" id="selectError">
<?php foreach($all as $nt) { echo "<option value =$nt[id]>$nt[name]</option>";}?>
</select>
</div>
回答by Nasser
Just Add an extra hidden option and print selected value from database
只需添加一个额外的隐藏选项并从数据库中打印选定的值
<option value="<?php echo $options;?>" hidden><?php echo $options;?></option>
<option value="PHP">PHP</option>
<option value="ASP">ASP</option>
回答by Pradeeshnarayan
This may help you.
这可能对你有帮助。
?php
$sql = "select * from mine where username = '$user' ";
$res = mysql_query($sql);
while($list = mysql_fetch_assoc($res))
{
$category = $list['category'];
$username = $list['username'];
$options = $list['options'];
?>
<input type="text" name="category" value="<?php echo '$category' ?>" readonly="readonly" />
<select name="course">
<option value="0">Please Select Option</option>
// Assuming $list['options'] is a coma seperated options string
$arr=explode(",",$list['options']);
<?php foreach ($arr as $value) { ?>
<option value="<?php echo $value; ?>"><?php echo $value; ?></option>
<?php } >
</select>
<?php
}
?>
