PHP和MySQL的下拉选项中选择问题

当我尝试从下拉列表中选择SQL选择的选项，我得到的所有选项中选择：PHP和MySQL的下拉选项中选择问题

$status1 = $row7["status"]; 

if($status1 == "draft") { $slc = "selected"; } 
if($status1 == "ordered") { $slc = "selected"; } 
if($status1 == "shipped") { $slc = "selected"; } 

echo " 
    <select class=\"form-control input-sm\" name = \"o_status\"> 
    <option value = \"draft\" $slc>draft</option> 
    <option value = \"ordered\" $slc>ordered</option> 
    <option value = \"shipped\" $slc>shipped</option> 
    </select> 
";

我与输出的问题选择了所有选项那样：

<select class="form-control input-sm" name = "status"> 
<option value = "draft" selected>draft</option> 
<option value = "ordered" selected>shipped</option> 
<option value = "shipped" selected>shipped</option>

来源

2017-04-04 Sky

请立即检查。

希望这有助于你：

$status1 = $row7["status"]; 

if($status1 == "draft") { $draft = "selected"; } 
if($status1 == "ordered") { $ordered = "selected"; } 
if($status1 == "shipped") { $shipped = "selected"; } 

    echo " 
    <select class=\"form-control input-sm\" name = \"o_status\"> 
<option value = \"draft\" $draft>draft</option> 
<option value = \"ordered\" $ordered>ordered</option> 
<option value = \"shipped\" $shipped>shipped</option> 
</select> 
    ";

来源

2017-04-04 11:04:00 Aaron

这将导致错误，如 PHP注意：未定义的变量：在第22行的thisfile.php中排序当选择其中一个时，其他未定义。看看我的回复;） –

Thx Ajay现在工作很好:) – Sky

欢迎:) – Aaron

您可以以设置选择选项定义状态的数组：

<?php 
    $statuses = array(
     'draft' => 'New draft', 
     'ordered' => 'New order', 
     'shipped' => 'New ship' 
    ); 
?> 

<select class="form-control input-sm" name="status"> 
<?php 
    foreach ($statuses as $status => $name) { 
?> 
    <option value="<?php echo $status ?>" <?php echo $status === $row7['status'] ? 'selected="selected"' : '' ?>> 
     <?php echo $name ?> 
    </option> 
<?php 
    } 
?> 
</select>

来源

2017-04-04 10:59:11

如果选项名称不同的期权价值如何设置呢？ – Sky

Like：等等。 – Sky

是的，您可以：试试这个： '<？php $ statuses = array（'draft'=>'something1'，'ordered'=> 'something2'，'shipped'=>'something3'）; $ row7 ['status'] ='草稿'; ？> <选择类= “形式控制输入-SM” 名称= “状态”> <？PHP的的foreach（$状态为$状态=> $值）{ ？> <选项值=” <？php echo $ value？>“<？php echo $ status === $ row7 ['status']？ '选择= “选择”'： '' >> <？PHP的echo $状态> ' –

也许这会有所帮助：

<?php 
switch ($row7["status"]) { 
    case 'draft': 
     $draft = "selected"; 
     $shipped = ""; 
     $ordered = ""; 
     break; 
    case 'ordered': 
     $ordered = "selected"; 
     $draft = ""; 
     $shipped = ""; 
     break; 
    case 'shipped': 
     $shipped = "selected"; 
     $draft = ""; 
     $ordered = ""; 
     break; 
    default: 
     $draft = ""; 
     $shipped = ""; 
     $ordered = ""; 
     break; 
} 
echo "<select class=\"form-control input-sm\" name = \"o_status\"> 
    <option value = \"draft\" $draft>draft</option> 
    <option value = \"ordered\" $ordered>ordered</option> 
    <option value = \"shipped\" $shipped>shipped</option> 
    </select>"; 
?>

来源

2017-04-04 11:08:34

PHP和MySQL的下拉选项中选择问题

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