我有一个小问题，我不知道为什么。也许有人可以帮助我。首先是一些免责声明;我还在学习PHP，我知道mysqli或pdo，但是这个服务器会运行一个旧版本的php。
现在好了解问题。
我有一个表单，它传递给我的post-data.php表单以推送到SQL db。
然而，当它推送数据时，它只是从表单的后期操作中推送变量而不是变量中的数据。
Screenshot of submitted data in PHPmyadmin
我的守则如下：
<?php
$hostname = "localhost"; $username = "goldme_owner";
$dbName = "goldme_dealer_meeting";
$connect = mysql_connect($hostname, $username);
if (!$connect) {
echo "Please try later.";
}
else {
mysql_select_db($dbName, $connect);
$checkboxA1 = isset($_POST['checkboxA1']) ? $_POST['checkboxA1'] : 'No';
$checkboxE1 = isset($_POST['checkboxE1']) ? $_POST['checkboxE1'] : 'No';
$checkboxF1 = isset($_POST['checkboxF1']) ? $_POST['checkboxF1'] : 'No';
$checkboxG1 = isset($_POST['checkboxG1']) ? $_POST['checkboxG1'] : 'No';
$checkboxH1 = isset($_POST['checkboxH1']) ? $_POST['checkboxH1'] : 'No';
$checkboxI1 = isset($_POST['checkboxI1']) ? $_POST['checkboxI1'] : 'No';
$checkboxJ1 = isset($_POST['checkboxJ1']) ? $_POST['checkboxJ1'] : 'No';
$checkboxK1 = isset($_POST['checkboxK1']) ? $_POST['checkboxK1'] : 'No';
}
echo "$checkboxA1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$checkboxE1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$checkboxF1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$checkboxG1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$checkboxH1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$checkboxI1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$checkboxJ1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$checkboxK1"; //just want to make sure checkbox vars are passing //will delete in final code
echo "$_POST[confirm]"; //just want to make sure confirm code generated //will delete in final code
$sql_statement = 'INSERT INTO 2014_registrations'.
'(confirm_number,timecode,company_name,country,address1,address2,city'.
',state,zip,phone,fax,email,zone_manager,transport,first_name,'.
'last_name,tuesday_tours,tuesday_meat,wed_lunch,wed_dinner,'.
'wed_pool_tourney,thurs_lunch,thurs_dinner,fri_shop,fri_tour,dietary)'.
'VALUES ("$_POST[confirm]","$_POST[timecode]","$_POST[company_name]",'.
'"$_POST[address]","$_POST[address2]","$_POST[city]","$_POST[state]",'.
'"$_POST[zip]","$_POST[country]","$_POST[phone]","$_POST[fax]",'.
'"$_POST[email]","$_POST[zonemanager]","$_POST[transport]",'.
'"$_POST[fattendee1]","$_POST[lattendee1]","$_POST[checkboxA1]",'.
'"$_POST[radio1]","$_POST[checkboxE1]","$_POST[checkboxF1]",'.
'"$_POST[checkboxG1]","$_POST[checkboxH1]","$_POST[checkboxI1]",'.
'"$_POST[checkboxJ1]","$_POST[checkboxK1]","$_POST[dietary1]")';
$rec_insert = mysql_query($sql_statement);
if(! $rec_insert ){
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($connect);
?>
答案 0 :(得分：0)
变量被读作字符串，因为它们用单引号书写。
我已将变量移到单引号之外，只是将它们连接到字符串。 此代码极易受SQL注入影响，不应在生产环境中使用。
$sql_statement = 'INSERT INTO 2014_registrations'.
'(confirm_number,timecode,company_name,country,address1,address2,city'.
',state,zip,phone,fax,email,zone_manager,transport,first_name,'.
'last_name,tuesday_tours,tuesday_meat,wed_lunch,wed_dinner,'.
'wed_pool_tourney,thurs_lunch,thurs_dinner,fri_shop,fri_tour,dietary)'.
'VALUES ("'.$_POST[confirm].'","'.$_POST[timecode].'","'.$_POST[company_name].'",'.
'"'.$_POST[address].'","'.$_POST[address2].'","'.$_POST[city].'","'.$_POST[state].'",'.
'"'.$_POST[zip].'","'.$_POST[country].'","'.$_POST[phone].'","'.$_POST[fax].'",'.
'"'.$_POST[email].'","'.$_POST[zonemanager].'","'.$_POST[transport].'",'.
'"'.$_POST[fattendee1].'","'.$_POST[lattendee1].'","'.$_POST[checkboxA1].'",'.
'"'.$_POST[radio1].'","'.$_POST[checkboxE1].'","'.$_POST[checkboxF1].'",'.
'"'.$_POST[checkboxG1].'","'.$_POST[checkboxH1].'","'.$_POST[checkboxI1].'",'.
'"'.$_POST[checkboxJ1].'","'.$_POST[checkboxK1].'","'.$_POST[dietary1].'")';
答案 1 :(得分：0)
首先，从POST请求直接将数据插入数据库几乎不是一个好主意。
我知道格式错误，但请尝试使用此$ sql_statement：
$sql_statement = 'INSERT INTO 2014_registrations'.
'(confirm_number,timecode,company_name,country,address1,address2,city'.
',state,zip,phone,fax,email,zone_manager,transport,first_name,'.
'last_name,tuesday_tours,tuesday_meat,wed_lunch,wed_dinner,'.
'wed_pool_tourney,thurs_lunch,thurs_dinner,fri_shop,fri_tour,dietary)'.
"VALUES (\"{$_POST[confirm]}\",\"{$_POST[timecode]}\",\"{$_POST[company_name]}\",".
"\"{$_POST[address]}\",\"{$_POST[address2]}\",\"{$_POST[city]}\",\"{$_POST[state]}\",".
"\"{$_POST[zip]}\",\"{$_POST[country]}\",\"{$_POST[phone]}\",\"{$_POST[fax]}\",".
"\"{$_POST[email]}\",\"{$_POST[zonemanager]}\",\"{$_POST[transport]}\",".
"\"{$_POST[fattendee1]}\",\"{$_POST[lattendee1]}\",\"{$_POST[checkboxA1]}\",".
"\"{$_POST[radio1]}\",\"{$_POST[checkboxE1]}\",\"{$_POST[checkboxF1]}\",".
"\"{$_POST[checkboxG1]}\",\"{$_POST[checkboxH1]}\",\"{$_POST[checkboxI1]}\",".
"\"{$_POST[checkboxJ1]}\",\"{$_POST[checkboxK1]}\",\"{$_POST[dietary1]}\")";
答案 2 :(得分：0)
你的`sql_statment错了。
$sql_statement = 'INSERT INTO 2014_registrations'.
'(confirm_number,timecode,company_name,country,address1,address2,city'.
',state,zip,phone,fax,email,zone_manager,transport,first_name,'.
'last_name,tuesday_tours,tuesday_meat,wed_lunch,wed_dinner,'.
'wed_pool_tourney,thurs_lunch,thurs_dinner,fri_shop,fri_tour,dietary)'.
'VALUES ("$_POST[confirm]","$_POST[timecode]","$_POST[company_name]",'.
'"$_POST[address]","$_POST[address2]","$_POST[city]","$_POST[state]",'.
'"$_POST[zip]","$_POST[country]","$_POST[phone]","$_POST[fax]",'.
'"$_POST[email]","$_POST[zonemanager]","$_POST[transport]",'.
'"$_POST[fattendee1]","$_POST[lattendee1]","$_POST[checkboxA1]",'.
'"$_POST[radio1]","$_POST[checkboxE1]","$_POST[checkboxF1]",'.
'"$_POST[checkboxG1]","$_POST[checkboxH1]","$_POST[checkboxI1]",'.
'"$_POST[checkboxJ1]","$_POST[checkboxK1]","$_POST[dietary1]")';
更改为：
$sql_statment = "INSERT INTO 2014_registrations".
"(confirm_number,timecode,company_name,country,address1,address2,city".
",state,zip,phone,fax,email,zone_manager,transport,first_name,".
"last_name,tuesday_tours,tuesday_meat,wed_lunch,wed_dinner,".
"wed_pool_tourney,thurs_lunch,thurs_dinner,fri_shop,fri_tour,dietary)".
"VALUES ($_POST['confirm'],$_POST['timecode'],$_POST['company_name'],".
"$_POST['address'],$_POST['address2'],$_POST['city'],$_POST['state'],".
"$_POST['zip'],$_POST['country],$_POST[phone'],$_POST['fax'],".
"$_POST['email'],$_POST['zonemanager'],$_POST['transport'],".
"$_POST['fattendee1'],$_POST['lattendee1'],$_POST['checkboxA1'],".
"$_POST['radio1'],$_POST['checkboxE1'],$_POST['checkboxF1'],".
"$_POST['checkboxG1'],$_POST['checkboxH1'],$_POST['checkboxI1'],".
"$_POST['checkboxJ1'],$_POST['checkboxK1'],$_POST['dietary1']")";
如果您使用$sql_statment = ''，则无法在$sql_statment = '$_POST[confirm]';中添加变量，因为它会将其视为文本。你必须像$sql_statment = $_POST['checkboxG1'].','.$_POST['checkboxH1'].','.$_POST['checkboxI1'].','.那样
另一种方法是像我一样将它全部更改为$sql_statment = "$_POST['checkboxG1'],$_POST['checkboxH1'],$_POST['checkboxI1'],";。
看到使用'和'时存在差异;
顺便说一下。你应该使用一些安全性，例如mysql_escape_string。
答案 3 :(得分：0)
在单引号内，变量不会被读取为变量。例如，echo '$a'将打印$a，但echo "$a"将打印$a的值。
在您的代码中，您尝试使用与'$a'类似的内容，换句话说，您使用围绕变量的单引号而不是双引号。
尝试将其写为
"VALUES ('$_POST[confirm]','$_POST[timecode]','$_POST[company_name]',".
或
'VALUES ("'.$_POST[confirm].'","'.$_POST[timecode].'","'.$_POST[company_name].'",'.
依旧......
请注意，使用$_POST[confirm]会发出通知。使用它的正确方法是$_POST['confirm']。
另请注意，您的代码容易受到SQL注入攻击。考虑使用准备好的陈述。