php 从 MySQL 获取值以检查单选按钮会生成通知
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Getting the value from MySQL to check radio buttons generates a notice
提问by SporeDev
I'm trying to set the "checked" status of a radio button based on the value from the MySQL database. The main purpose of the code is to be a "edit user profile data" script. I managed to get all information from the database and everything gets edited as it should if the user choses to but I can't seem to make the radio buttons work. I get the following message:
我正在尝试根据 MySQL 数据库中的值设置单选按钮的“已检查”状态。代码的主要目的是成为“编辑用户配置文件数据”脚本。我设法从数据库中获取了所有信息，并且如果用户选择的话，一切都会按原样进行编辑，但我似乎无法使单选按钮起作用。我收到以下消息：
Notice: Use of undefined constant Male - assumed 'Male' in C:\xampp\htdocs\Signup\user_panel.php on line 57
注意：使用未定义的常量男性 - 在 C:\xampp\htdocs\Signup\user_panel.php 中第 57 行假定为“男性”
I checked SO for this notice but I couldn't find anything related to radio buttons AND the notice.
我为此通知检查了 SO，但找不到与单选按钮和通知相关的任何内容。
This is the code:
这是代码：
<form action="includes/user_panel.php" method="POST">
<table>
<tr>
<td><label for="email"> E-mail </label></td>
<td><input type="email" name="email" id="email" value="<?php echo "{$fetch['email']}"; ?>"></td>
</tr>
<tr>
<td><label for="first_name"> First name </label></td>
<td><input type="text" name="first_name" id="first_name" value="<?php echo "{$fetch['first_name']}"; ?>"></td>
</tr>
<tr>
<td><label for="last_name"> Last name </label></td>
<td><input type="text" name="last_name" id="last_name" value="<?php echo "{$fetch['last_name']}"; ?>"></td>
</tr>
<tr>
<td><label for="birthday"> Birthday </label></td>
<td><input type="date" name="birthday" id="birthday" value="<?php echo "{$fetch['birthday']}"; ?>"></td>
</tr>
<tr>
<td><label for="sex"> Sex </label></td>
<?php
if ($fetch['sex'] = male)
{
echo "<td><input type='radio' name='sex' value='male' id='sex' checked> Male <input type='radio' name='sex' value='female' id='sex'> Female </td>";
}
else
{
echo "<td><input type='radio' name='sex' value='male' id='sex'> Male <input type='radio' name='sex' value='female' id='sex' checked> Female </td>";
}
?>
</tr>
</table>
</br>
<input type="submit" name="edit_user" value="Edit" class="button_1">
<input type="hidden" name="edit_user_data">
</form>
Additional information:
附加信息：
$initial_email = $_SESSION['email'];
$username = $_SESSION['username'];
$query_user_panel="SELECT * FROM users WHERE email = '$initial_email' OR username= '$username'";
$result_user_panel=mysql_query($query_user_panel) or die (mysql_error());
$fetch = mysql_fetch_assoc($result_user_panel)
Line 57 is the one with the "if".
第 57 行是带有“if”的那一行。
Scroll down the code and you'll see where the sex should get checked. I tried with a "if" but that generates me the notice that I wrote.
向下滚动代码，您将看到应该检查性别的位置。我尝试使用“如果”，但这会生成我写的通知。
I know that MySQL is depreciated and I should use MySQLi. If you have the time please check this question as well: Simplest MySQL to MySQLi transition
我知道 MySQL 已折旧，我应该使用 MySQLi。如果您有时间，也请检查这个问题：Simplest MySQL to MySQLi transition
Thank you very much and please let me know if I need to edit my post or provide with additional data.
非常感谢，如果我需要编辑我的帖子或提供其他数据，请告诉我。
回答by Vilsol
Your problem is that you have to forgot to put == instead of = as well as you have to put Male in quotes.
你的问题是你必须忘记把 == 而不是 = 以及你必须把 Male 放在引号中。
It should look like this:
它应该是这样的：
if ($fetch['birthday'] == "Male")
Also, are you sure that birthday field is correct one to check for sex?
另外，您确定生日字段是检查性别的正确字段吗？
回答by Shemeer M Ali
I found there are three mistakes
我发现有三个错误
1) $fetch['birthday'], I think it should be another field name like sex
1）$fetch['birthday']，我觉得应该是另外一个字段名，比如sex
2) if ($fetch['birthday'] = Male), =should be ==so, if ($fetch['birthday'] == "Male")
2) if ($fetch['birthday'] = Male),=应该是==这样,if ($fetch['birthday'] == "Male")
3) if ($fetch['birthday'] == "Male"), you missed double quotes for string so should be "Male"
3) if ($fetch['birthday'] == "Male"), 你错过了字符串的双引号所以应该是 "Male"
<?php
if ($fetch['birthday'] == "Male")
{
echo "<td><input type='radio' name='sex' value='male' id='sex' checked> Male <input type='radio' name='sex' value='female' id='sex'> Female </td>";
}
else
{
echo "<td><input type='radio' name='sex' value='male' id='sex'> Male <input type='radio' name='sex' value='female' id='sex' checked> Female </td>";
}
?>
回答by Allan Starr
I also wanted a solution where when data is pulled from the database to edit a form, if the account is "Active" it will check the "Active" radio button and if the account is "Inactive" it will check the Inactiveradio button on the html form. I have used tinyint in the mysql database using: 1 - Activeand 0 - InactiveThe code i have used that does this successfully is below:
我也想一个解决方案，当数据从数据库中抽取到编辑形式，如果该帐户是“激活”它会检查“活动”单选按钮，如果账户是“无效”，它会检查无效单选按钮html表单。我在 mysql 数据库中使用了 tinyint：1 - Active和 0 - Inactive我使用的成功执行此操作的代码如下：
<label for="Active">Active</label>
<input id="Active" type="radio" name="Active" value="1"<?php if ($row['Active'] == 1): ?> checked = "checked"<?php endif; ?> />
<label for="Active">Inactive</label>
<input id="Active" type="radio" name="Active" value="0"<?php if ($row['Active'] != 1): ?> checked = "checked"<?php endif; ?> />
So when i query a record and it has a value of '1' in the database field then the Activeradio button will be filled and the Inactivebutton empty and when i query a record and it has a value of '0' in the database field then the Inactiveradio button will be filled and the Activebutton empty.
因此，当我查询一条记录并且它在数据库字段中的值为“ 1”时，活动单选按钮将被填充，而“非活动”按钮为空；当我查询一条记录时，它在数据库中的值为“ 0”字段，然后将填充Inactive单选按钮，而Active按钮为空。
Hope this helps
希望这可以帮助
回答by Ahmed Yousry Ebrahim
try this if ($fetch['sex'] == 'Male')
试试这个 if ($fetch['sex'] == 'Male')
