从MYSQL,PHP更新数据下拉按钮

时间:2022-09-25 23:25:00
if(isset($_GET["id"])){
        $sql=mysql_query("SELECT * FROM aMovie WHERE aName= '{$_GET['id']}'"); 
        $row=mysql_fetch_object($sql);  
}

<input type = "text" name = "name" value = "<?php echo $row->aC; ?>"/> 
<select name = "name" >
        <option value = "" <?php echo ($row->aC== "Deadpool") ? 'selected = "selected"': '';?>">Deadpool</option>
        <option value = "" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected = "selected"': '';?>">BATMAN VS SUPERMAN</option>
</select>

Assume that aMovie is my table name, and in my table there are aName and and aC. However, I would want to display aName which matches aC ["Deadpool" or "Batman Vs Superman"] and display it in the drop down button. It only works for the input type but not the drop down button.

假设aMovie是我的表名,在我的表中有aName和aC。但是,我想显示一个匹配aC [“Deadpool”或“Batman Vs Superman”]的名称,并在下拉按钮中显示。它仅适用于输入类型,但不适用于下拉按钮。

4 个解决方案

#1


3  

Your <select> should be like:

你的

<select name = "name" >
<option value="Deadpool" <?=($rows->aC == "Deadpool" ? 'selected="selected"': '')?>>Deadpool</option>
<option value="BATMAN VS SUPERMAN" <?=($rows->aC == "BATMAN VS SUPERMAN" ? 'selected="selected"': '')?>>BATMAN VS SUPERMAN</option>
</select>

selected="selected" will use outside the value attribute.

selected =“selected”将在value属性之外使用。

UPDATE:

更新:

As @Maninderpreet-Singh mentioned, you also need to change $row to $rows.

正如@ Maninderpreet-Singh所提到的,你还需要将$ row更改为$ rows。

#2


1  

try to change

试着改变

<option <?php echo($row->aC== "Deadpool") ? 'selected = "selected"': '';?>  value="<?php echo $row->aC;?>">Deadpool</option>

#3


1  

try with this and you are using different variable in input

试试这个,你在输入中使用不同的变量

<input type = "text" name = "name" value = "<?php echo $rows->aC; ?>"/> 

$row and $rows are different

$ row和$ rows是不同的

  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "Deadpool") ? 'selected':'';?>">Deadpool</option>
  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected': '';?>">BATMAN VS SUPERMAN</option>

#4


1  

<input type = "text" name = "name1" value = "<?php echo $rows->aC; ?>"/> 
<select name = "name2" >
        <option value = " <?php echo($row->aC); ?>" <?php echo($row->aC=="Deadpool")?'selected': '';?>>Deadpool</option>
        <option value = " <?php echo($row->aC);?>" <?php echo($row->aC == "BATMAN VS SUPERMAN")?'selected': '';?>>BATMAN VS SUPERMAN</option>
</select>

#1


3  

Your <select> should be like:

你的

<select name = "name" >
<option value="Deadpool" <?=($rows->aC == "Deadpool" ? 'selected="selected"': '')?>>Deadpool</option>
<option value="BATMAN VS SUPERMAN" <?=($rows->aC == "BATMAN VS SUPERMAN" ? 'selected="selected"': '')?>>BATMAN VS SUPERMAN</option>
</select>

selected="selected" will use outside the value attribute.

selected =“selected”将在value属性之外使用。

UPDATE:

更新:

As @Maninderpreet-Singh mentioned, you also need to change $row to $rows.

正如@ Maninderpreet-Singh所提到的,你还需要将$ row更改为$ rows。

#2


1  

try to change

试着改变

<option <?php echo($row->aC== "Deadpool") ? 'selected = "selected"': '';?>  value="<?php echo $row->aC;?>">Deadpool</option>

#3


1  

try with this and you are using different variable in input

试试这个,你在输入中使用不同的变量

<input type = "text" name = "name" value = "<?php echo $rows->aC; ?>"/> 

$row and $rows are different

$ row和$ rows是不同的

  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "Deadpool") ? 'selected':'';?>">Deadpool</option>
  <option value = "<?php echo $row->aC; ?>" <?php echo ($row->aC == "BATMAN VS SUPERMAN") ? 'selected': '';?>">BATMAN VS SUPERMAN</option>

#4


1  

<input type = "text" name = "name1" value = "<?php echo $rows->aC; ?>"/> 
<select name = "name2" >
        <option value = " <?php echo($row->aC); ?>" <?php echo($row->aC=="Deadpool")?'selected': '';?>>Deadpool</option>
        <option value = " <?php echo($row->aC);?>" <?php echo($row->aC == "BATMAN VS SUPERMAN")?'selected': '';?>>BATMAN VS SUPERMAN</option>
</select>