[英]PHP/MSSQL Present stored field value in a drop-down list
I use the following code to create a number of drop-down lists with the same values. 我使用以下代码创建多个具有相同值的下拉列表。 The values are brought in from an MSSQL table via a query written elsewhere. 这些值是通过在其他地方编写的查询从MSSQL表中引入的。
<?php
$select = '';
while($row = $data->fetch(PDO::FETCH_BOTH))
{
$select .= "<option value='".$row['Code']."'>".$row['Code']."</option> ";
}
echo "<select name=\"proj1[]\">";
echo $select;
echo "</select>";
?>
The user makes his selections, then submits the form and the record is written to the PROJECTS table in the PROJ DB (fields: Proj1, Proj2, Proj3, Proj4). 用户进行选择,然后提交表单,并将记录写入PROJ DB(字段:Proj1,Proj2,Proj3,Proj4)中的PROJECTS表中。 The original drop-down values are held in a separate table (CODES). 原始下拉值保存在单独的表(CODES)中。 When the record is called up in a browser, a prepared SELECT statement runs against PROJECTS to load it. 在浏览器中调用记录时,针对PROJECTS运行准备好的SELECT语句以加载它。 I would like to show the user the drop-down selections he made when the completed form is loaded, ie, the values of Proj1-Proj4 for a given record in PROJECTS. 我想向用户展示他在加载完成的表单时所做的下拉选择,即PROJECTS中给定记录的Proj1-Proj4值。 How can I do this? 我怎样才能做到这一点? I'm not sure where to put my 'option selected'. 我不确定将“选择的选项”放在哪里。
Hope this help 希望这个帮助
$select = '';
while($row = $data->fetch(PDO::FETCH_BOTH))
{
$select .= "<option value='".$row['Code']."' '".$row['Code'] == $_POST['proj1'] ? ' selected="selected"' : ''."' >".$row['Code']."</option> ";
}
echo "<select name=\"proj1[]\">";
echo $select;
echo "</select>";
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