PHP求助table获取错误

Question

可以帮我解决这个问题吗？我只想替换单行的值。我的意思是，我有两个 tables:Subjects 和 Careers ->

"Subjects"包括(id,careers_id(是table职业的"id"列的外键) 主题、描述、时间）

"Careers" 包括 (id,name,description)

我只想替换外键的值，它将为我们提供职业名称

最后我需要这样的东西：

型号

这是我的代码：

<html>
<head>





</head>
<body>
<a href="estudiante.php">ADD NEW SUBJECT</a><br /><br />
    <h2 align="center">TABLE:SUBJECTS</h2>
        <table align="center" border="1" cellspacing="0" cellpadding="0" width="700">
            <thead>
                <th>id</th>
                <th>Career</th>
                <th>Subject</th>
                <th>Description</th>
                <th>Hours</th>
                <th>Action</th>

            </thead>

            <?php
            $sql = mysql_query("SELECT * FROM subjects");
            $i=1;
                while($row = mysql_fetch_array($sql)){
                    echo "<tr>
                            <td>".$i."</td>
                            <td>".$row['careers_id']."</td>
                            <td>".$row['subject']."</td>
                            <td>".$row['description']."</td>
                            <td>".$row['hours']."</td>
                            <td align='center'>
                                <a href='editar.php?editar=1&iden=".$row['id']."'>UPDATE</a> |
                                <a href='borrar.php?borrar=1&iden=".$row['id']."'>DELETE</a>
                            </td>
                    </tr>";
                    $i++;

                }
            ?>

        </table>    

</body>

而且，我放了一个按钮，可以让我添加一个新主题。因此，当我单击它时，另一个页面 open.I 需要添加一个 slider/select，它向我显示 table 职业中可用的职业。看一看，我需要这样的东西：

SELECT/SLIDER

这是我添加新主题的代码（但我不知道如何制作 slider/select :/）

<?php include('connect.php'); 
$error="";

if(isset($_POST['btnsave']))
{
    $carreras_id=$_POST['txtcarreras_id'];
    $subject=$_POST['txtsubject'];
    $descripcion=$_POST['txtdescripcion'];
    $carga_horaria=$_POST['txtcarga_horaria'];


    if($_POST['txtid']=="0")
    {

        $a_sql=mysql_query("INSERT INTO subjects VALUES('','$carreras_id','$subject','$descripcion','$carga_horaria')");
        if($a_sql)
        {

            header("location:index.php");

        }


    }else{

        echo "Actualizar";
    }

}



?>

        <h2 align="center">ADD NEW SUBJECT</h2>
        <form method="Post">
            <table align="center">
                <tr>    
                    <td>Career:</td>
                    <td><input type='text' name='txtcarreras_id'/><input type="hidden" name="txtid" value="0" /></td>

                </tr>
                <tr>    
                    <td>Subject:</td>
                    <td><input type='text' name='txtsubject'/></td>

                </tr>
                <tr>    
                    <td>Description:</td>
                    <td><input type='text' name='txtdescripcion'/></td>

                </tr>
                <tr>    
                    <td>Hours:</td>
                    <td><input type='text' name='txtcarga_horaria'/></td>

                </tr>
                <tr>    
                    <td></td>
                    <td><input type='submit' value=save name='btnsave'/></td>

                </tr>
            </table>


        </form>

希望你能帮助我:/

谢谢！

Answer 1

您需要在查询中加入职业table：

SELECT s.*, c.name AS career FROM subjects s LEFT JOIN careers c ON s.careers_id = c.id.

之后，您可以使用$row['career']访问职业名称。

Answer 2

（抱歉英语不好）

<?php include('connect.php'); 
$error="";

if(isset($_POST['btnsave']))
{
    $carreras_id=$_POST['txtcarreras_id'];
    $subject=$_POST['txtsubject'];
    $descripcion=$_POST['txtdescripcion'];
    $carga_horaria=$_POST['txtcarga_horaria'];

    if($_POST['txtid']=="0")
    {

        $a_sql=mysql_query("INSERT INTO subjects VALUES('','$carreras_id','$subject','$descripcion','$carga_horaria')");
        if($a_sql)
        {

            header("location:index.php");

        }


    }else{

        echo "Actualizar";
    }

}
//NEW CODE for get careers list START HERE!
//do this outside your 'post' check because
//you need $careers array available for populate your form's select field
$careers = array();


$c_sql=mysql_query("SELECT * FROM careers"); //get careers data

while($row = mysql_fetch_assoc($c_sql))
{
    $careers[] = $row; // store each career in $careers variable
}
//NEW CODE for get careers list END HERE!
?>



<h2 align="center">ADD NEW SUBJECT</h2>

<form method="post"> /*post in lowercase*/
    <table align="center">
        <tr>    
            <td>Career:</td>
            <td>
              /*NEW CODE for show careers list START HERE!*/
              <select name='txtcarreras_id'>
              <?php
              foreach($careers as $career)
              {
                  echo "<option value='{$career['id']}'>{$career['name']}</option>";
              }
              ?>
              </select>
              /*NEW CODE for show careers list END HERE!*/
              <input type="hidden" name="txtid" value="0" />
            </td>
        </tr>
        <tr>    
            <td>Subject:</td>
            <td><input type='text' name='txtsubject'/></td>
        </tr>
        <tr>    
            <td>Description:</td>
            <td><input type='text' name='txtdescripcion'/></td>
        </tr>
        <tr>    
            <td>Hours:</td>
            <td><input type='text' name='txtcarga_horaria'/></td>
        </tr>
        <tr>    
            <td></td>
            <td><input type='submit' value=save name='btnsave'/></td>
        </tr>
    </table>
</form>

注意：我建议使用mysqli（过程或面向对象）或Pdo而不是mysql，因为mysql_ * 函数将被弃用。

EXTRA: 我假设 connect.php 看起来类似于：

connect.php:

<?php

$conn = mysql_connect($server, $user, $pass);

if (!$conn) {
    echo "Can't connect to mysql:".mysql_error();
    exit;
}

if (!mysql_select_db($dbname)) {
    echo "Can't select Database {$dbname}: " . mysql_error();
    exit;
}

使用 mysqli 对象将是：

connect.php:

<?php

function getDB()
{
    $conn = new mysqli($server, $user, $pass);

    if ($conn->connect_errno) {
        echo "Can't connect to mysql:".$conn->connect_error;
        exit;
    }

    $conn->select_db($dbname);

    if ($conn->connect_errno) {
        echo "Can't select Database {$dbname}: " . $conn->connect_error;
        exit;
    }
    return $conn;
}
//getDB() will return your connection object

您可以这样查询：

<?php
include('connect.php'); 
//$sql = "SELECT * FROM sometable";
//$resource = getDB()->query($sql);
//now resource will have the result of your query

//for select your careers
$c_sql = getDB()->query("SELECT * FROM careers"); //get careers data

//while($row = mysql_fetch_assoc($c_sql))
//{
//    $careers[] = $row; // store each career in $careers variable
//}

//LESS CODE!
$careers = $c_sql->fetch_all(MYSQLI_ASSOC) // MYSQLI_ASSOC for associative array

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