PHP和MySQL的下拉選項中選擇問題

當我嘗試從下拉列表中選擇SQL選擇的選項，我得到的所有選項中選擇：PHP和MySQL的下拉選項中選擇問題

$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> 
";

我與輸出的問題選擇了所有選項那樣：

<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>

來源

2017-04-04 Sky

請立即檢查。

希望這有助於你：

$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> 
    ";

來源

2017-04-04 11:04:00 Aaron

這將導致錯誤，如 PHP注意：未定義的變量：在第22行的thisfile.php中排序當選擇其中一個時，其他未定義。看看我的回覆;） –

Thx Ajay現在工作很好:) – Sky

歡迎:) – Aaron

您可以以設置選擇選項定義狀態的數組：

<?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>

來源

2017-04-04 10:59:11

如果選項名稱不同的期權價值如何設置呢？ – Sky

Like：等等。 – Sky

是的，您可以：試試這個： '<？php $ statuses = array（'draft'=>'something1'，'ordered'=> 'something2'，'shipped'=>'something3'）; $ row7 ['status'] ='草稿'; ？> <選擇類= 「形式控制輸入-SM」名稱= 「狀態」> <？PHP的的foreach（$狀態爲$狀態=> $值）{ ？> <選項值=」 <？php echo $ value？>「<？php echo $ status === $ row7 ['status']？ '選擇= 「選擇」'： '' >> <？PHP的echo $狀態> ' –

也許這會有所幫助：

<?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>"; 
?>

來源

2017-04-04 11:08:34

PHP和MySQL的下拉選項中選擇問題

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