在 PHP 和 MySQL 中重新打开页面时，在下拉列表中显示保存的数据 - display saved data in dropdown list when reopen the page in php and mysql

display saved data in dropdown list when reopen the page in php and mysql

当我再次登录时，是否可以在从mysql到php的下拉列表中显示保存的值？我的代码将值保存在 Mysql 中。但是当我重新登录时，它会显示空白字段（下拉列表中默认）

$sql = "SELECT Program_Description FROM programs";
$result = mysqli_query($dbc, $sql);
$dropdown = "<select name='Program_Description'>";
$dropdown .= "<option value= ></option>";
while($row = mysqli_fetch_assoc($result)) {
$dropdown .= "'r'n<option value='{$row['Program_Description']}'>{$row ['Program_Description']}     </option>";
}
$dropdowna .= "'r'n</select>";

这样做

 //Suppose the $selecteValue contains the selected value which you have fetched from the database
 $selectedValue = 'test';
 $sql = "SELECT Program_Description FROM programs";
 $result = mysqli_query($dbc, $sql);
 $dropdown = "<select name='Program_Description'>";
 $dropdown .= "<option value= ></option>";
 while($row = mysqli_fetch_assoc($result)) {
      $dropdown .= "'r'n<option ".(($selectedValue == $row['Program_Description']) ? ' selected ' : '')." value='{$row['Program_Description']}'>{$row['Program_Description']}     </option>";
 }
 $dropdowna .= "'r'n</select>";

另一种解决方案是

   $sql = "SELECT Program_Description FROM programs";
   $result = mysqli_query($dbc, $sql);
   $dropdown = "<select name='Program_Description'>";
   $dropdown .= "<option value= ></option>";
   $strSelect = '';
   while($row = mysqli_fetch_assoc($result)) {
         if ($selectedValue == $row['Program_Description']) {   $strSelect = ' selected '; } else { $strSelect =  ''; }
        $dropdown .= "'r'n<option ".$strSelect." value='{$row['Program_Description']}'>{$row['Program_Description']}     </option>";
   }
   $dropdowna .= "'r'n</select>";

试试这个，

    $sql = "SELECT Program_Description FROM programs";
$result = mysqli_query($dbc, $sql);
$savedValueInDb = ""; //fetch the DB saved value and assign to this variable
$dropdown = "<select name='Program_Description'>";
$dropdown .= "<option value= ></option>";
while($row = mysqli_fetch_assoc($result)) {
$dropdown .= "'r'n<option value='{$row['Program_Description']}'";
$dropdown .= ($savedValueInDb == $row ['Program_Description']) ? " selected " : "";
$dropdown .= ">{$row ['Program_Description']}     </option>";
}
$dropdowna .= "'r'n</select>";