表单连接和更新到数据库

时间：2016-03-22 07:59:57

标签： php mysql

我无法更新数据库中的数据，但它显示数据是成功的关键，我也有几个错误，但我不知道如何纠正它们，因为我＆＃39 ;新来的php..please帮助我。

我的firstform.php：

<?php
session_start();
include 'connect.php';
$query = mysql_query("SELECT *FROM firstform WHERE $id='$_SESSION[manager_name]'");

while($row = mysql_fetch_assoc($query))
{
    $manager =$row['manager_name'];
    $evaluator =$row['evaluator_name'];
    $outlet =$row['outlet_location'];
    $date =$row['date'];
    $day =$row['day'];
    $timein =$row['time_in'];
    $timeout =$row['time_out'];
    $overall=$row['overall'];
    $quality=$row['quality'];
    $service=$row['service'];
    $cleanliness=$row['cleanliness'];
}
?>

这是我的updatefirstform.php：

<?php
//error_reporting(0);
include'connect.php';
session_start();

$manager =$_POST['manager_name'];
$evaluator =$_POST['evaluator_name'];
$outlet =$_POST['outlet_location'];
$date =$_POST['date'];
$day =$_POST['day'];
$timein =$_POST['time_in'];
$timeout =$_POST['time_out'];
$overall=$_POST['overall'];
$quality=$_POST['quality'];
$service=$_POST['service'];
$cleanliness=$_POST['cleanliness'];

mysql_query("UPDATE firstform SET id ='$id' manager_name='$manager', evaluator_name ='$evaluator',outlet_location='$outlet',date='$date',day='$day',time_in ='$timein',time_out='$timeout',overall='$overall',quality='$quality',service='$service',cleanliness='$cleanliness' WHERE $id ='$_SESSION[manager_name]'");

echo ("<SCRIPT LANGUAGE ='JavaScript'>
         window.alert ('Data are Sucessfully Updated!')
         window.location.href ='secondform.php'
         </SCRIPT>");

?>

错误：

Errors :Notice: Undefined variable: manager in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 4

Notice: Undefined index: manager_name in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 4

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 6

4 个答案:

答案 0 :(得分：1)

设置ID后忘记了,。在更新条件中，为什么$id而不是id？顺便提一下$id的价值是多少？它应该是这样的：

mysql_query("UPDATE firstform SET id ='$id',
                                  manager_name='$manager',
                                  evaluator_name ='$evaluator',
                                  outlet_location='$outlet',
                                  date='$date',
                                  day='$day',
                                  time_in ='$timein',
                                  time_out='$timeout',
                                  overall='$overall',
                                  quality='$quality',
                                  service='$service',
                                  cleanliness='$cleanliness'
                              WHERE id = '$_SESSION[manager_name]'");

与SELECT语句相同：

$query = mysql_query("SELECT * FROM firstform WHERE id = '".$_SESSION['manager_name']."'");

在将变量绑定到查询之前，也请使用*_real_escape_string。

$manager = mysql_real_escape_string($_POST['manager_name']);
//         ^^ do this for the rest of your passed-on variables

您还应该考虑使用mysqli_*而不是弃用的mysql_*。

答案 1 :(得分：0)

Private Sub Timer1__Tick()...
    call GET_QUEUED_EMAILS
    Timer2.Start
    Timer1.Stop
End sub
Private Sub Timer2__Tick()...
    call Function2
    Timer3.Start
    Timer2.Stop
End sub
Private Sub Timer3__Tick()...
    call Function3
    Timer3.Start
End sub

更新

中的相同问题

答案 2 :(得分：0)

首先，manager_name发布数据不能来自您的更新表单，请检查它。

之后使用它

    <?xml version="1.0" encoding="UTF-8"?>
    <project xmlns="http://maven.apache.org/POM/4.0.0"
     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
     xsi:schemaLocation="http://maven.apache.org/POM/4.0.0   http://maven.apache.org/xsd/maven-4.0.0.xsd">
   <modelVersion>4.0.0</modelVersion>

 <groupId>JERSEYProject</groupId>
 <artifactId>JERSEYProject</artifactId>
 <packaging>war</packaging>
<version>1.0-SNAPSHOT</version>
<name>simple-service-webapp</name>

<build>
    <finalName>simple-service-webapp</finalName>
    <plugins>
        <plugin>
            <groupId>org.apache.maven.plugins</groupId>
            <artifactId>maven-compiler-plugin</artifactId>
            <version>2.5.1</version>
            <inherited>true</inherited>
            <configuration>
                <source>1.7</source>
                <target>1.7</target>
            </configuration>
        </plugin>
    </plugins>
</build>


<dependencyManagement>
    <dependencies>
        <dependency>
            <groupId>org.glassfish.jersey</groupId>
            <artifactId>jersey-bom</artifactId>
            <version>${jersey.version}</version>
            <type>pom</type>
            <scope>import</scope>
        </dependency>
    </dependencies>
</dependencyManagement>

<dependencies>
<dependency>
    <groupId>org.glassfish.jersey.containers</groupId>
    <artifactId>jersey-container-servlet-core</artifactId>
</dependency>
</dependencies>

答案 3 :(得分：0)

我编辑了我的firstform.php：

<?php
//error_reporting(0);
include'connect.php';
session_start();
if(isset($_POST['submit'])) {


    $id=mysqli_real_escape_string($_POST['user_id']);
    $manager =mysqli_real_escape_string($_POST['manager_name']);
    $evaluator =mysqli_real_escape_string($_POST['evaluator_name']);
    $outlet =mysqli_real_escape_string($_POST['outlet_location']);
    $date =mysqli_real_escape_string($_POST['date']);
    $day =mysqli_real_escape_string($_POST['day']);
    $timein =mysqli_real_escape_string($_POST['time_in']);
    $timeout =mysqli_real_escape_string($_POST['time_out']);
    $overall=mysqli_real_escape_string($_POST['overall']);
    $quality=mysqli_real_escape_string($_POST['quality']);
    $service=mysqli_real_escape_string($_POST['service']);
    $cleanliness=mysqli_real_escape_string($_POST['cleanliness']);

mysqli_query("UPDATE firstform SET manager_name='$manager', evaluator_name ='$evaluator',outlet_location='$outlet',date='$date',day='$day',time_in ='$timein',time_out='$timeout',overall='$overall',quality='$quality',service='$service',cleanliness='$cleanliness' WHERE $id ='$_SESSION[user_id]'");
}
if (isset($_GET['submit'])) {

echo ("<SCRIPT LANGUAGE ='JavaScript'>
         window.alert ('Data are Sucessfully Updated!')
         window.location.href ='secondform.php'
         </SCRIPT>");
}
?>

<?php
mysql_close($connection);
?>

and i'm getting this error again, how to fix it?


    Notice: Undefined variable: id in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 5

    Notice: Undefined index: user_id in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 5

    Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\marrybrown_clean\firstform.php on line 6

also each my data in my form shows undefined variable like :

    <br /><b>Notice</b>:  Undefined variable: manager in <b>C:\xampp\htdocs\marrybrown_clean\firstform.php</b> on line <b>139</b><br />

i think my form is okay, but why is it showing error? if i take the id away, how to start my session by manager name only?

    This would be my form :

    <h2><font size ="5">Pre Evaluation Checklist</font></h2>
                        <form action="updatefirstform.php" method="post">
                            <div class="form_settings">
                                <p><input class ="input" type ="hidden" name="user_id" value ="<?php echo $id; ?>" /> </p>
                                <p><span>Manager on Duty  :</span><input type="manager" id="manager_name" name="manager_name" value="<?php echo $manager;?>" /></p>
                                <p><span>Evaluator's Name :</span><input type="evaluator_name" name="evaluator_name" value="<?php echo $evaluator;?>" /></p>
                                <p><span>Outlet Location  :</span><input type="outlet_location" name="outlet_location" value="<?php echo $outlet;?>"/></p>
                                <p><span>Date             :</span><input type="date" name="date" value="<?php echo $date;?>"  /></p>
                                <p><span>Day              :</span><input type="day" name="day" value="<?php echo $day;?>"  /></p>
                                <p><span>Time In          :</span><input type="timein" name="time_in" value="<?php echo $timein;?>" /></p>
                                <p><span>Time Out         :</span><input type="timeout" name="time_out" value="<?php echo $timeout;?>" /></p>
                                <table width="100%" height="79" border-spacing:0;>
                        <tr><th width="15%">Overall </th><th width="15%">Quality</th><th width="15%">Service </th><th width="15%"> Cleanliness </th></tr>
                        <tr><td><input type="overall" name="overall" value ="<?php echo $overall;?>"/></td><td><input type="overall" name="quality" value="<?php echo$quality;?>"/> </td><td><input type="service" name="service" value ="<?php echo $service;?>"/></td><td><input type="cleanliness" name="cleanliness" value="<?php echo $cleanliness;?>"/></td></tr>

                        </table>

还有我的updatefirstform.php：

    #!/opt/lampp/bin/php
    <?php


    $host = "localhost";
    $port = 3027;
    $con = 1;

    set_time_limit(0);

    $socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP) or die('Could not create socketn');

    $result = socket_bind($socket, $host, $port) or die('Could not bind to socketn');

    $result = socket_listen($socket, 3) or die('Could not set up socket listenern');

    while (true)
    {
        $spawn = socket_accept($socket) or die('Could not accept incoming connectionn');

        if(socket_getpeername($spawn , $host , $port))
            {
            echo "Client $address : $port is now connected to us. \n";
            }


        $input = socket_read($spawn, 1024) or die('Could not read inputn');


        $input = trim($input);

        $output = strrev($input) . 'n';

        socket_write($spawn, $output, strlen ($output)) or die('Could not write output n');

    }

        socket_close($spawn);
        socket_close($socket);

    ?>
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