提交PHP表单不会将PHP变量传递给javascript函数
在我继承的php应用程序中,我是一个php新手,我有一个包含多个include的php表单。其中一个include设置了两个表,每个表中都有一个表单。第一个表单显示从MySQL数据库检索的记录,因此可能返回1个或多个记录。while循环遍历记录并用每个记录名、电话、电子邮件的数据填充控件。控件名为[fieldname]——其中$line从1开始,随着while循环遍历记录而递增。这很好用 当有人想要编辑其中一个字段并提交更改时,就会出现问题。表单有一个onsubmit=returnvalidateForm;我已经检查过以确保$line变量确实递增,但是当它被发送到javascript函数validateForm时,该变量没有在javasscript函数中定义。最初我没有传递$line变量,但javascript函数一直告诉我它无法获取未定义元素的值 第一格式代码提交PHP表单不会将PHP变量传递给javascript函数,javascript,php,forms,Javascript,Php,Forms,在我继承的php应用程序中,我是一个php新手,我有一个包含多个include的php表单。其中一个include设置了两个表,每个表中都有一个表单。第一个表单显示从MySQL数据库检索的记录,因此可能返回1个或多个记录。while循环遍历记录并用每个记录名、电话、电子邮件的数据填充控件。控件名为[fieldname]——其中$line从1开始,随着while循环遍历记录而递增。这很好用 当有人想要编辑其中一个字段并提交更改时,就会出现问题。表单有一个onsubmit=returnvalidat
<form action="admin_profiles_main_update.php" method="post" name="submit_order" enctype="multipart/form-data" onsubmit="return validateForm(<?php echo $line; ?>);">
<table border="0" cellspacing="0" cellpadding="0" class="forms">
<col width="20%"/>
<col width="80%"/>
<?php
if($user_access == 'National'){
$result = mysql_query("SELECT * FROM Profile ORDER BY profile_name");
}else{
$result = mysql_query("SELECT * FROM Profile WHERE profile_parent_region = '$user_profile' ORDER BY profile_name");
}
$line = 1;
while ($row_profile = mysql_fetch_array($result)){
$field_id = "_" . $line;
?>
<!-- LINE -->
<tr><td colspan="11"><hr class="view_line"></td></tr>
<!-- LINE -->
<tbody id="region<?php echo $field_id; ?>" >
<input class="field_long" name="profile_id<?php echo $field_id; ?>" id="profile_id<?php echo $field_id; ?>" type="hidden" value="<?php echo $row_profile[profile_id]; ?>"/>
<tr>
<td>
<label class="field_label" for="profile_parent_region<?php echo $field_id; ?>">Profile: </label>
</td>
<td align="left">
<select name="profile_parent_region<?php echo $field_id; ?>" id="profile_parent_region<?php echo $field_id; ?>" class="drop_med" >
<option></option>
<?php
if($user_access != 'National'){
$result_profile = mysql_query("Select Distinct profile_parent_region FROM profile where profile_parent_region = '$user_profile' ");
}else {
$result_profile = mysql_query("Select Distinct profile_parent_region FROM profile ORDER BY profile_parent_region ASC");
}
while($row = mysql_fetch_array($result_profile)){
echo '<option value ="'.$row['profile_parent_region'].'"';
if($row['profile_parent_region'] == $user_profile){
echo ' selected="selected"';
}
echo ' > ' . $row['profile_parent_region'] . '</option>';
}
?>
</select>
<span class="must_fill">* </span>
<label class="form_des" for="profile_parent_region<?php echo $field_id; ?>"></label>
</td>
</tr>
<tr>
<td>
<label class="field_label" for="profile_name<?php echo $field_id; ?>">Region: </label>
</td>
<td align="left">
<select name="profile_name<?php echo $field_id; ?>" id="profile_name<?php echo $field_id; ?>" class="drop_med" >
<option></option>
<?php
$parent_region = $row_profile['profile_name'];
if($user_access != 'National'){
$result_region = mysql_query("Select Distinct region FROM regions where parent_region = '$user_profile' ");
}else {
$result_region = mysql_query("Select Distinct region FROM regions ORDER BY region ASC");
}
while($row = mysql_fetch_array($result_region)){
echo '<option value ="'.$row['region'].'"';
if ($row['region'] == $parent_region){
echo ' selected="selected"';
}
echo ' >' . $row['region'] . '</option>';
}
?>
</select>
<span class="must_fill">* </span>
<label class="form_des" for="profile_name<?php echo $field_id; ?>"></label>
</td>
</tr>
<tr>
<td><label class="field_label" for="profile_manager<?php echo $field_id; ?>" >Region's Manager's Name: </label></td>
<td>
<input class="field_long" name="profile_manager<?php echo $field_id; ?>" id="profile_manager<?php echo $field_id; ?>" type="input" value="<?php echo $row_profile[profile_manager]; ?>"/>
<span class="must_fill">*</span>
<label class="form_des" for="profile_manager<?php echo $field_id; ?>"></label>
</td>
</tr>
<tr>
<td><label class="field_label" for="profile_phone<?php echo $field_id; ?>" >Region's Contact Number: </label></td>
<td>
<input class="field_long" name="profile_phone<?php echo $field_id; ?>" id="profile_phone<?php echo $field_id; ?>" type="input" value="<?php echo $row_profile[profile_phone]; ?>"/>
<span class="must_fill">*</span>
<label class="form_des" for="profile_phone<?php echo $field_id; ?>"></label>
</td>
</tr>
<tr>
<td><label class="field_label" for="profile_email<?php echo $field_id; ?>" >Region's Contact E-mail: </label></td>
<td>
<input class="field_long" name="profile_email<?php echo $field_id; ?>" id="profile_email<?php echo $field_id; ?>" type="input" value="<?php echo $row_profile[profile_email]; ?>"/>
<span class="must_fill">*</span>
<label class="form_des" for="profile_email">This email address will also be used to advise of files added to a submitted order.</label>
</td>
</tr>
<tr align="center">
<td colspan="2" class="loginrow">
<br />
<input name="Login <?php echo $field_id; ?>" id="Login<?php echo $field_id; ?>" value="Update Profile" type="submit" class="submit_button"/>
<br />
<br />
</td>
</tr>
</tbody>
<?php
$line++;
}
?>
</table>
</form>`
你的代码
不要使用mysql函数,因为它们已被弃用。使用mysqli或PDO。就这个javascript函数而言,暂时假设PHP不涉及。直接查看生成的HTML来诊断发生了什么。感谢TechBrat的建议,不幸的是生成的HTML真的没有给我任何帮助。感谢dognose,我使用了你的解决方案的一部分将表单放入循环,这很有帮助,但完全取消了javascript验证,因为它仍然不起作用,改为对提交php表单进行验证。谢谢回复!!!
if (lineNum == null) {
var x = document.forms["submit_order"]["file_name"].value;
if (x == null || x == '') {
alert("Missing File Name.");
return false;
}
var x = document.forms["submit_order"]["file_for"].value;
if (x == null || x == '') {
alert("Missing File Type.");
return false;
}
var x = document.forms["submit_order"]["OrderForm"].value;
if (x == null || x == '') {
alert("Missing File to Upload.");
return false;
}
}
<?php
$line = 0;
while ($row_profile = mysql_fetch_array($result)){
$field_id = "_" . $line;
?>
<form action="admin_profiles_main_update.php" method="post" name="submit_order" enctype="multipart/form-data" onsubmit="return validateForm(<?php echo $line; ?>);">
<!-- do all the row stuff -->
</form>
<?$line++; }?>