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我建立了一个将 Service Mechanic 客户存储到数据库的站点。它具有更新条目的功能。我有一个下拉选择字段工作正常（它仍然可以）。但是我在事实之后添加的第二个仅返回空白以获取信息。

这是我的代码。

include('connect.php');

$table='currentJobs';
$tableMap=array('Date','Name','Phone','Bike_Year','Bike_Model','Current_Status','SType','Mechanic','Revenue','Notes');

$sqlArray=array();

foreach ($tableMap AS $key){
    $sqlArray[]="`".$key."`='".mysql_real_escape_string(@$_POST[$key],$dbLink)."'";
}

$sql="UPDATE `".$table."` SET ".join(',',$sqlArray)." WHERE `id`='".$_POST['id']."'";

if (mysql_query($sql,$dbLink)){
    echo 'Data for <i>' .$_POST['Name'].'</i> was successfully updated <br />'.$sql.'.';
}
else {
    echo 'Sorry, could not process the following sql: <br /><code>'.$sql.'</code>';
    echo mysql_errno($dbLink) . ": " . mysql_error($dbLink). "\n";
}

mysql_close($dbLink);

它正在返回：

Data for Test User was successfully updated 
UPDATE `currentJobs` SET `Date`='04/14/2013',`Name`='Test User',`Phone`='1234567890',`Bike_Year`='2001',`Bike_Model`='FXD',`Current_Status`='Checked In',`SType`='',`Mechanic`='All',`Revenue`='0',`Notes`='' WHERE `id`='55'.

我的表格如下所示：

<form action="updated.php" method="POST" name="dataForm" id="dataForm">
<fieldset>
    <input type="hidden" name="id" value="<?php echo $row['id']; ?>"/>
    <label>Arrival<input type="text" name="Date" value="<?php echo $row['Date']; ?>"/></label><br />
    <label>Name<input type="text" name="Name" value="<?php echo $row['Name']; ?>"/></label><br />
    <label>Phone Number<input type="text" name="Phone" value="<?php echo $row['Phone']; ?>"/></label><br />
    <label>Bike Year<input type="text" name="Bike_Year" value="<?php echo $row['Bike_Year']; ?>"/></label><br />
    <label>Bike Model<input type="text" name="Bike_Model" value="<?php echo $row['Bike_Model']; ?>"/></label><br />
    <label>Current Status<!--Update Status types on two pages-->
        <select name="Current_Status" form="dataForm" value="<?php echo $row['Current_Status']; ?>">
            <option value="Checked In">Checked In</option>
            <option value="Inline For Service">Inline For Service</option>
            <option value="In Service">In Service</option>
            <option value="On Hold - Parts on Order">On Hold - Parts on Order</option>
            <option value="On Hold - Parts to Paint">On Hold - Parts to Paint</option>
            <option value="On Hold - Waiting To Hear From Customer">On Hold - Waiting To Hear From Customer</option>
            <option value="Test Ride">Test Ride</option>
            <option value="Completed - Awaiting Pick Up">Completed - Awaiting Pick Up</option>
            <option value="Picked Up">Picked Up</option>
        </select>
    </label><br />
    <!--update Service Type in two pages.-->
    <label>Service Type</label>
        <select name="SType" form="" value="<?php echo $row['SType']; ?>">
            <option value="Spec Service Interval">Spec Service Interval</option>
            <option value="Interim Service">Interim Service</option>
            <option value="Diagnostics">Diagnostics</option>
            <option value="Tires">Tires</option>
            <option value="Engine– Light Work (1-3 hrs)">Engine– Light Work (1-3 hrs)</option>
            <option value="Engine– Medium Work (3-8 hrs)">Engine– Medium Work (3-8 hrs)</option>
            <option value="Engine– Heavy Work (8-24 hrs)">Engine– Heavy Work (8-24 hrs)</option>
            <option value="Drivetrain– Light">Drivetrain– Light</option>
            <option value="Drivetrain - Heavy">Drivetrain - Heavy</option>
            <option value="Dyno Tune">Dyno Tune</option>
        </select>
    <br />
    <label>Mechanic Assigned<input type="text" name="Mechanic" value="<?php echo $row['Mechanic']; ?>"/></label><br />
    <label>Final Revenue<input type="text" name="Revenue" value="<?php echo $row['Revenue']; ?>"/></label><br />
    <label>Notes<input type="text" name="Notes" value="<?php echo $row['Notes']; ?>"/></label><br />
    <input type="submit" value="Update" style="color:#000000;" />
</fieldset>

为什么第二个选择不发送数据？

4

1 回答 1

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该select标签没有名为value.

http://www.w3schools.com/tags/tag_select.asp

更新：

像这样的东西会更有用：

<label>Service Type</label>
    <select name="SType" form="dataForm">
    <?php
    $Stypes = array(
    "Spec Service Interval",
    "Interim Service",
    "Diagnostics",
    "Tires",
    "Engine– Light Work (1-3 hrs)",
    "Engine– Medium Work (3-8 hrs)",
    "Engine– Heavy Work (8-24 hrs)",
    "Drivetrain– Light",
    "Drivetrain - Heavy",
    "Dyno Tune",
    );

    foreach($Stypes as $value){
      if ($value==$row['SType']){
        echo "<option value=\"$value\" selected=\"selected\">value</option>";  
      } else {
        echo "<option value=\"$value\">$value</option>";  
      }
    }

    ?>

    </select>

假设您$row['Stype']要从数据库中撤出。

同样正如@Fred 所说，如果您使用的是 HTML5，您可能需要为该属性指定一个表单 IDform（尽管这在 IE 中不起作用......但是）。

您还可以print_r($_POST)在进行插入的页面上进行调试，以验证发生了什么。

于 2013-04-14T20:40:33.543 回答