WordPress - 在特定页面上选择查询

时间:2016-01-07 17:59:13

标签: php mysql wordpress

我想在特定页面中选择查询。我在数据库中创建了表:

enter image description here

首先,我在插件上添加了phpexec以便在页面上使用PHP。然后,我测试了Select查询,这没关系。最后,我想创建用于检查序列号的表单。这是我的代码:

<html>
<head>
    <title></title>
</head>
<body>
    <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST"><br><br>   
    Serial Number: <input type="Text" Name="Num1"><p>
    <input type="Submit" value="Calculate">
    </form>
<phpcode>
 <?php  
        if (count($_POST) > 0 && isset($_POST["Num1"])
        {
            $servername = "localhost";
            $username = "******";
            $password = "*******";
            $dbname = "******";
            $serialNum = $_POST["Num1"];

            // Create connection
            $conn = new mysqli($servername, $username, $password, $dbname);
            // Check connection
            if ($conn->connect_error) {
                die("Connection failed: " . $conn->connect_error);
            } 

            $sql = "SELECT Install FROM SN WHERE serial = $serialNum";
            $result = $conn->query($sql);

            if ($result->num_rows > 0) {

                while($row = $result->fetch_assoc()) {
                    echo "id: " . $row["Install"];
                }
            } else {
                echo "no result";
            }
            $conn->close();
        }
 ?> 
</phpcode>
</body>
</html>

实际输出为:

enter image description here

如何解决这个问题?

2 个答案:

答案 0 :(得分:0)

你的表格坏了......试试这种方式

class City:

def __init__(self):
    self.coordinates = set()

def gift(self, x, y):
    self.coordinates.add((x, y))

def was_gifted(self, x, y):
    return (x, y) in self.coordinates # Does this return True or False?????

def all_gifted(self):
    return self.coordinates


class Santa(City):

def __init__(self, City):
    self.city = City
    self.coords = (0, 0)

def move(self, c):
    if c == "^":
        self.coords[1] -= 1
    elif c == "ˇ":
        self.coords[1] += 1
    elif c == "<":
        self.coords[0] -= 1
    elif c == ">":
        self.coords[0] += 1

def moves(self, path):
    for i in path:
        self.move(i)

def gift(self):
    self.city.gift

class FastSanta:
?????????????

答案 1 :(得分:0)

嗨,也许你可以试试这个 是一个小例子,检查你的值和id,值的名称 和sintaxis for html和php

<html>
<head>
    <title></title>
</head>
<body>
    <form name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
    <br><br>  
    Serial Number: <input type="text" name="Num1" id="Num1">
    <br><br>  
    <input type="submit" name="submit" id="submit" value="Calculate">
    <br><br>  
    </form>
<phpcode>
 <?php  
   if(isset($_POST["submit"])) 
   {
           function Conect()
        {
            if (!($link=mysql_connect("localhost","username","password")))
            {
                echo "error to conect to database.";
                exit();
            }
            if (!mysql_select_db("databasename",$link))
            {
                echo "Error to select database.";
                exit();
            }
            return $link;
        }//end function Conect
        $serialNum = mysql_real_escape_string($_POST["Num1"]);
        $query="SELECT Install FROM SN WHERE serial = $serialNum";
        $action=mysql_query($query,$link) or die("Error: ".mysql_error());

        if(mysql_num_rows($action) > 0)
        {
            ?>
            <table border="1">
               <tr COLSPAN=2 BGCOLOR="#6D8FFF">
                  <td>INSTALL</td>
               </tr>
             <?php     
             while($row=mysql_fetch_array($action))
             {
              echo "<tr>".
                   "<td>".$row["Install"]."</td>".
                   "</tr>";
             }//end while 

        }
        else
        {
             echo "don't exist recordsfor list ";
        }//end if
        mysql_close($link);

    }//end if
 ?> 
</phpcode>
</body>
</html>
祝你好运.. !!