我目前有一个编辑表单,其中所有值都从db回显为每个输入的“值”,然后用户将重新键入并“保存”该值以附加到db。所有回声都有效,但我无法回应选项的数据库值...任何人都知道如何?
&安培;感谢
以下代码:
<fieldset>
<form id="input-form" method="POST" action="../php/edit-import-record.php">
Airway Bill Number*:
<input type=text name=AwbNo size=30 class="input" value="<?php echo $awb ?>" required>
Client Code:
<?php //OPEN DROP DOWN BOX
include("../login/dbinfo.inc.php");
$comm=@mysql_connect(localhost,$username,$password);
$rs=@mysql_select_db($database) or die( "Unable to select database");
$sql= "SELECT DISTINCT ClientCode, ClientName FROM tbl_client ORDER BY ClientCode";
$result = mysql_query($sql);
echo "<select name='ClientCode'>";
while ($row = mysql_fetch_array($result)) {
echo "<option value='" . $row['ClientCode'] ."'>" . $row['ClientName'] ."</option>";
}
echo "</select><br>"; //CLOSE DROP DOWN BOX
?><br>
Vessel Name*:
<input type=text name=VesselName class=form-control id=inputSuccess size=30 class="input" value="<?php echo $vsl ?>" required>
Number of Pieces:
<input type=number name=Pieces size=30 class="input" value="<?php echo $pcs ?>" >
Total Weight (kg):
<input type=number name=Weight size=30 class="input" value="<?php echo $wgt ?>" >
Carrier:
<input type=text name=Carrier size=30 class="input" value="<?php echo $car ?>" >
Sender:
<input type=text name=Sender size=30 class="input" value="<?php echo $snd ?>" >
Status:
<input type=text name=Status size=30 class="input" value="<?php echo $stt ?>" >
Arrival Date:
<input type=date name=ArrivalDate size=30 class="input" value="<?php echo $ard ?>" >
Customs:
<input type=text name=Customs size=30 class="input" value="<?php $ctm ?>" >
<br><small>Fields marked with * are required to be filled in.</small>
<div class="inputformbutton">
<button type="reset" class="btn btn-default btn-sm">Reset</button>
<button type="submit" class="btn btn-primary btn-sm">Create Record</button>
</div>
</fieldset>