比较$ _GET []中的值并选择一个选项

Question

我需要在$ _GET []给出的某些条件下选择“选中”的解决方案。 这个想法是。

<?
$sql_id="select * from stock order by id asc";
$result_id=mysql_db_query($dbname,$sql_id);
while($rec_id=mysql_fetch-array($result_id)){
  $_id=$rec_id['id'];
  $_title=$rec_id['title'];
if($_GET['id']==$_id){
  echo "<option value=\"$_id\" selected>$_title</option>";
}else{//I need this option selected where id='2'
  echo "<option value=\"$_id\">$_title</option>";
}
?>

问题是：我还需要空（$ _ GET ['id']）= 2。那么条件就是

1. 当$ _GET ['id'] == $ _ id
时选择一个选项
2. 当$ _GET ['id'] ==“”
时，从mysql（其中stock.id ='2'）中选择一个选项

Answer 1

不确定我是否理解正确，但你为什么不尝试这样的事情：

if($_GET['id']==$_id){
  echo "<option value=\"$_id\" selected>$_title</option>";
}
else if (($_GET['id']=="") && ($_id==2)){//I need this option selected where id='2'
  echo "<option value=\"$_id\">$_title</option>";
}

Answer 2

<?php
    $query = "SELECT * from `".$dbname."`.`stock` ORDER BY `id` ASC";
    $result_id = mysql_query($query);
    $id = isset($_GET['id']) ? (int)$_GET['id'] : 0;
    while($rec_id = mysql_fetch_array($result_id)){
        $sel = '';
        if($rec_id['id'] == $id){
            $sel = " selected='selected'";
        }
        if(!$id && $rec_id['id'] == 2){
            $sel = " selected='selected'";
        }
        echo "<option value='".$rec_id['id']."'".$sel.">".$rec_id['title']."</option>";
    }
?>

Answer 3

我现在知道了......

<?
//let's test the $_GET['id'] value first
if(!$_GET['id']){
    $_id="2";//if no $_GET['id'] then $_id==2
}elseif(isset($_GET['id'])){
    $_id=$_GET['id'];
}

$sql_id="select * from stock order by id asc";
$result_id=mysql_db_query($dbname,$sql_id);
while($rec_id=mysql_fetch-array($result_id)){
  $_id=$rec_id['id'];
  $_title=$rec_id['title'];
if($_GET['id']==$_id){
  echo "<option value=\"$_id\" selected>$_title</option>";
}else{//I need this option selected where id='2'
  echo "<option value=\"$_id\">$_title</option>";
}
?>

问题解决了！非常感谢你的帮助！