Php MySql结果显示的是id号,而不是单词
我有一个表格,我填写,一旦提交它将返回成功的消息与您的提交。 除了我不知道如何获得成功的消息来显示给定变量的单词外,其他一切都工作得很好。例如: 在我提交一个条目后,你会得到这样一个输出 你电话的详细情况 会计问题 (2) 来自Michael,关于a(1) 采取的行动: 会计-更改密码。 这个问题解决了吗?(一) 我想说的是 帐户问题 (来电)来自Michael,关于(密码重置) 采取的行动: 会计-更改密码 这个问题解决了吗?(是的) 我试着做一个左连接,但我不确定我是否做对了。 这是我的密码-Php MySql结果显示的是id号,而不是单词,php,mysql,join,Php,Mysql,Join,我有一个表格,我填写,一旦提交它将返回成功的消息与您的提交。 除了我不知道如何获得成功的消息来显示给定变量的单词外,其他一切都工作得很好。例如: 在我提交一个条目后,你会得到这样一个输出 你电话的详细情况 会计问题 (2) 来自Michael,关于a(1) 采取的行动: 会计-更改密码。 这个问题解决了吗?(一) 我想说的是 帐户问题 (来电)来自Michael,关于(密码重置) 采取的行动: 会计-更改密码 这个问题解决了吗?(是的) 我试着做一个左连接,但我不确定我是否做对了。 这是我的密码
if(isset($_POST['new_support'])) {
$tech_id = $_POST['tech_id'];
$callform_id = $_POST['callform_id'];
$issues_id = $_POST['issues_id'];
$cbr = $_POST['cbr'];
$custname = $_POST['custname'];
$connection_id = $_POST['connection_id'];
$os_id = $_POST['os_id'];
$device_id = $_POST['device_id'];
$email_id = $_POST['email_id'];
$comments = $_POST['comments'];
$resolved_id = $_POST['resolved_id'];
$address_id = $_POST['address_id'];
$date = time();
if($error == '') {
$sql = "INSERT INTO support (tech_id, callform_id, issues_id, account_id, connection_id, router_id, connectedto_id, os_id, device_id, wired_id, email_id, hosting_id, customer_id, resolved_id, address_id, cbr, custname, comments)
VALUES ('$tech_id', '$callform_id', '$issues_id', '$account_id', '$connection_id', '$router_id', '$connectedto_id', '$os_id', '$device_id', '$wired_id', '$email_id', '$hosting_id', '$customer_id', '$resolved_id', '$address_id', '$cbr', '$custname', '$comments')";
$query = mysql_query($sql) or die("Snilbogs ".mysql_error());
$sql2 = "SELECT * FROM support LEFT JOIN callform ON support.callform_id = callform.id LEFT JOIN issues ON support.issues_id = issues.id LEFT JOIN account ON support.account_id = account.id LEFT JOIN connection ON support.connection_id = connection.id LEFT JOIN router ON support.router_id = router.id LEFT JOIN connectedto ON support.connectedto_id = connectedto.id LEFT JOIN os ON support.os_id = os.id LEFT JOIN device ON support.device_id = device.id LEFT JOIN wired ON support.wired_id = wired.id LEFT JOIN email ON support.email_id = email.id LEFT JOIN hosting ON support.hosting_id = hosting.id LEFT JOIN resolved ON support.resolved_id = resolved.id WHERE cbr=$cbr";
$result = mysql_query($sql2) or die ("Snilbogs something went wrong here".mysql_error());
while($row = mysql_fetch_array($result)); {
echo $row['callform_id'];
echo "<h2>Success!</h2>";
echo "<div class='success_message'>Thank you! </div>";
echo "<h2>Details of your call</h2>";
echo "<ul class='success-reg'>";
if($issues_id == '1'){
echo "<li><span class='success-info'><b>Accounting Issues</b></span><br />
($callform_id) from $custname, regarding a $account_id<br />
Actions Taken: <br />
$comments.<br />
This issus is Resolved? $resolved_id <br />";
}
}
if(!isset($_POST['new_support']) || $error != '') {
echo $error;
if(isset($\u POST['new\u support'])){
$tech\u id=$\u POST['tech\u id'];
$callform\u id=$\u POST['callform\u id'];
$issues\u id=$\u POST['issues\u id'];
$cbr=$_POST['cbr'];
$custname=$_POST['custname'];
$connection\u id=$\u POST['connection\u id'];
$os_id=$_POST['os_id'];
$device\u id=$\u POST['device\u id'];
$email\u id=$\u POST['email\u id'];
$comments=$_POST['comments'];
$resolved_id=$_POST['resolved_id'];
$address\u id=$\u POST['address\u id'];
$date=time();
如果($error=''){
$sql=“插入到支持中(技术id、呼叫表单id、问题id、帐户id、连接id、路由器id、连接到id、操作系统id、设备id、有线id、电子邮件id、主机id、客户id、已解决id、地址id、cbr、客户名称、评论)
值(“$tech_id”、“$callform_id”、“$issues_id”、“$account_id”、“$connection_id”、“$router_id”、“$connectedto_id”、“$os_id”、“$device_id”、“$wired_id”、“$email_id”、“$hosting_id”、“$customer_id”、“$resolved_id”、“$resolved($address_id”、“$address($cbr”、“$custname”、$comments”);
$query=mysql\u query($sql)或die(“Snilbogs.mysql\u error());
$sql2="选择*从support上的support LEFT JOIN callform.callform\u id=callform.id支持上的LEFT JOIN问题。问题\u id=issues.id支持上的LEFT JOIN帐户。帐户\u id=account.id支持上的LEFT JOIN连接。连接\u id=connection.id支持上的LEFT JOIN路由器。路由器\u id=router.id支持上的LEFT JOIN连接。连接到\u id=connectedto.id LEFT JOIN os ON support.os_id=os.id LEFT JOIN device ON support.device_id=device.id LEFT JOIN wired ON support.wired_id=wired.id LEFT JOIN email ON support.email_id=email.id LEFT JOIN hosting ON support.hosting_id=hosting.id LEFT JOIN resolved ON support.resolved_id=resolved.id其中cbr=$cbr”;
$result=mysql\u query($sql2)或die(“Snilbogs这里出了点问题”。mysql\u error());
而($row=mysql_fetch_array($result)){
echo$row['callform_id'];
呼应“成功!”;
回声:“谢谢!”;
回显“通话详情”;
echo“”;
如果($issues\u id==“1”){
echo“- 会计问题
$custname中的($callform_id),关于$account_id
采取的措施:
$comments.
此问题已解决?$Resolved_id
;
}
}
如果(!isset($_POST['new_support'])| |$error!=''){
echo$错误;
这很简单,您只需请求表支持中的内容。您必须在SELECT中指定还需要来自联接表的列。尝试使用print\u r()打印mysql\u fetch\u数组($result)时会得到什么?您使用的是发布的值($\u POST
),而不是数据库查询中的值($row
)。