php form_dropdown 中的 CodeIgniter select_value
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CodeIgniter select_value in form_dropdown
提问by user1269625
I have a dropdown field in my form and when I hit submit the form goes through validation and if an error happens, return all the value. This work expect for my dropdown meu, I have the set_value in the dropdown but it doesnt work :(
我的表单中有一个下拉字段，当我点击提交时，表单会通过验证，如果发生错误，则返回所有值。这项工作期待我的下拉菜单，我在下拉菜单中有 set_value 但它不起作用:(
Here is my code
这是我的代码
<?php echo form_dropdown('gender', $gender, set_value('gender')); ?>
What am I doing wrong or missing?
我做错了什么或错过了什么？
回答by user1269625
Doing this worked well:
这样做效果很好：
<?php
$selected = ($this->input->post('gender')) ? $this->input->post('gender') : 'M';
$gender = array("M" => "Male", "F" => "Female");
echo form_dropdown('gender', $gender, $selected);
?>
回答by brightintro
Remove the set_value('gender')
去除那个 set_value('gender')
As in:
如：
<?php echo form_dropdown('gender', $gender, 'male'); ?>
As Trung was mentioning you can pass an array as the 3rd parameter for multiple selections.
正如 Trung 所提到的，您可以将数组作为多个选择的第三个参数传递。
回答by NULL pointer
If you want to use set_value for a specific field, you need to have a validation rule for that field. No validation rule makes set_value return NULL (or an empty string I have not checked which)
如果要对特定字段使用 set_value，则需要为该字段设置验证规则。没有验证规则使 set_value 返回 NULL（或我没有检查过的空字符串）
In my controller I added
在我的控制器中，我添加了
$this->form_validation->set_rules('gender','Gender','required');
before my form_validation->run line. I could then used your syntax:
在我的 form_validation->run 行之前。然后我可以使用你的语法：
<?php echo form_dropdown('gender', $gender, set_value('gender')); ?>
I found set_value() worked, NOT set_select(). Remember you can add a second parameter to set_value to have a default of male or female which is used if the form has not yet been validated. I extract the existing value from my backend database, and pass this in as such:
我发现 set_value() 有效，而不是 set_select()。请记住，您可以将第二个参数添加到 set_value 以具有默认的男性或女性，如果表单尚未验证，则使用该参数。我从我的后端数据库中提取现有值，并将其传递如下：
<?php echo form_dropdown('gender', $gender, set_value('gender',$persons_gender_from_db)); ?>
Note also that set value returns the index of the array of options('M' or 'F'), not the displayed value ('Male' or 'Female'). For lists containing more options that two genders, I use the primary key of the database table containing the options, to ensure uniqueness.
另请注意，set value 返回选项数组的索引（'M' 或 'F'），而不是显示值（'Male' 或 'Female'）。对于包含两个性别的更多选项的列表，我使用包含选项的数据库表的主键，以确保唯一性。
Although I wonder when I will first be asked to add 'Other' to the list of choices for gender....
虽然我想知道什么时候我会第一次被要求将“其他”添加到性别选择列表中......
回答by Hieu Le
With form_dropdown, you have to use set_selectinstead of set_value
使用form_dropdown，您必须使用set_select代替set_value
If you want to set the default value on form_dropdown, pass an array which contains defaults value as the third parameter to the form_dropdownfunction.
如果要在 上设置默认值form_dropdown，请将包含默认值的数组作为第三个参数传递给form_dropdown函数。
回答by pbarney
CodeIgniter has set_checkbox()and set_radio()that you need to use instead of set_value()
CodeIgniter 具有set_checkbox()并且set_radio()您需要使用而不是set_value()
But set_checkboxand set_radiohave some issues, and don't seem to be able to handle forms that are designed to handle BOTH create AND update forms.
但是，set_checkbox和set_radio有一些问题，并且似乎没有能够处理被设计为同时处理创建和更新表单形式。
This is the fix. You can put this code in a helper or extend the form_validation class.
这是修复。您可以将此代码放在帮助程序中或扩展 form_validation 类。
<?php echo form_dropdown('gender', $gender, set_it('gender','M',$person)); ?>
<?php
/* $field is the field you're setting
* $value is the "selected" value from the previous form post
* $defaults is an object containing defaults in case the form is being used to create a new record. It could be filled with null values, or actual default values if you need that.
*/
function set_it($field, $value, $defaults = null)
{
// first, check to see if the form element was POSTed
if (isset($_POST[$field]))
{
// does it match the value we provided?
if ($_POST[$field] == $value)
{
// yes, so set the checkbox
echo "checked='checked'"; // valid for both checkboxes and radio buttons
}
}
// There was no POST, so check to see if the provided defaults contains our field
elseif ( ! is_null($defaults) && isset($defaults->$field))
{
// does it match the value we provided?
if ($defaults->$field == $value)
{
// yes, so set the checkbox
echo "checked='checked'"; // valid for both checkboxes and radio buttons
}
}
}
?>
回答by Yuri Giovani
Probably you are not validating the form.
可能您没有验证表单。
Use this:
用这个：
$this-> form_validation-> set_rules ('gender', 'Label', 'xss_clean');
To use:
使用：
<? php echo form_dropdown ('gender', $ gender, set_value ('gender'));?>
If not please use the form validation, do as follows:
如果不是，请使用表单验证，请执行以下操作：
<? php echo form_dropdown ('gender', $ gender, $ this-> input-> post ('gender'));?>
