我试图在检查到 mysql 数据库表时插入单选按钮的值。下面是这样做的 HTML 和 PHP。请让我知道出了什么问题?
Here is the HTML first:
<div class='container'>
<label for='username' >Business*:</label><br/>
<input type="radio" name="bus" id="username" value="bus" maxlength="50" /><br/>
<span id='register_username_errorloc' class='error'></span>
</div>
<div class='container'>
<label for='username' >Personal*:</label><br/>
<input type="radio" name="pers" id="username" value="per" maxlength="50" /><br/>
<span id='register_username_errorloc' class='error'></span>
</div>
Now the PHP:
function InsertIntoDB(&$formvars)
{
$confirmcode = $this->MakeConfirmationMd5($formvars['email']);
$formvars['confirmcode'] = $confirmcode;
$insert_query = 'insert into '.$this->tablename.'(
name,
email,
username,
password,
confirmcode,
dob,
business,
personal
)
values
(
"' . $this->SanitizeForSQL($formvars['name']) . '",
"' . $this->SanitizeForSQL($formvars['email']) . '",
"' . $this->SanitizeForSQL($formvars['username']) . '",
"' . md5($formvars['password']) . '",
"' . $confirmcode . '",
"' . $this->SanitizeForSQL($formvars['dob']) . '",
"' . $this->SanitizeForSQL($formvars['bus']) . '",
"' . $this->SanitizeForSQL($formvars['pers']) . '"
)';
if(!mysql_query( $insert_query ,$this->connection))
{
$this->HandleDBError("Error inserting data to the table\nquery:$insert_query");
return false;
}
return true;
}
最佳答案
我觉得有逻辑错误。您应该只在数据库中保存单选按钮中的变量之一。
"' . $this->SanitizeForSQL($formvars['bus']) . '", "' . $this->SanitizeForSQL($formvars['pers']) . '"
关于php - 插入单选按钮的值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/14781907/