Question

我正在建立一个网站来学习编码。我有一个claim.php，这是一个声明您的业务页面（它只是一个搜索业务的地方，然后单击addclaimedbiz.php的链接）和一个addclaimedbiz.php页面，它将声明的业务添加到数据库。

首先，继承我的用户的SQL代码，其中业务ID将添加到行 biz ：

`id` int(15) NOT NULL AUTO_INCREMENT,
`firstname` varchar(50) NOT NULL,
`lastname` varchar(50) NOT NULL,
`email` varchar(75) NOT NULL,
`password` varchar(50) NOT NULL,
`biz` int(1) NOT NULL,
`verifiedbiz` int(1) NOT NULL,

并且是我的企业的SQL代码：

`id` int(15) NOT NULL AUTO_INCREMENT,
`name` varchar(50) NOT NULL,
`phone` varchar(14) NOT NULL,
`claimed` int(1) NOT NULL,

现在继承了claim.php的代码，该代码将公司ID发送到addclaimedbiz.php。（我只发了一点，但如果你需要它可以发布更多！）

 while($row = mysql_fetch_array($result))
            {
            $id=$row['id'];
            $company_name=$row['name'];
            $company_phone=$row['phone'];
            $company_address=$row['address'];
            $address2=$row['address2'];
            $company_city=$row['city'];
            $company_zip=$row['zipcode'];
            $cat1=$row['cat1'];
            $cat2=$row['cat2'];
            $cat3=$row['cat3'];
            $subcat1=$row['subcat1'];
            $subcat2=$row['subcat2'];
            $subcat3=$row['subcat3'];
            $claimed=$row['claimed'];
            //Start While Loop
            echo"
            <div class='listing'>
                                    <li>
                                        <span class='bphone'>$company_phone</span>
                                        <span class='bname'>$company_name</span>
                                        <br/>
                                        <div class='blocation'>$company_address, $company_city, CO $company_zip ";if($address2 != ""){echo"(".$address2.")";}echo"</div>
                                        <br/>
                                        ";if($claimed != 1){echo "<a href='addclaimedbiz.php?id=".$id."'><button>Claim Business</button></a>";}else{echo "Already Claimed";}echo"
                                    </li>

                                <!--/Listing-->

                </div>";

            }

然后是我的addclaimedbiz.php的代码：

 <?
 $biz_id = $_REQUEST['id'];

 //This next select from db is to take the id of the company and get the name for the corresponding id

 include("./config.php");
 $result = mysql_query("SELECT * FROM company WHERE id = '$biz_id'") or die(mysql_error());
 while($row = mysql_fetch_array($result))
 {
    $business_name = $row['name'];
 }

 ?>

然后回显我使用的$ business_name（这是在addclaimedbiz.php以及页面下方的一点点）

<?php if($biz == "0"){

            echo "You are claiming:"$business_name " as your business";

            }
            else{"You are already have a claimed business"}
            ?>

应该发生的事情是它应该回应你声称：$ business_name作为你的业务，但现在它遗漏了$ business_name。这是为什么？

主要感谢所有人的帮助！

哦，我遗漏了数据库中的一些行，这些行对我想到的问题是不必要的，但如果你需要它们，我绝对可以发帖！

修改

我的新代码如下所示：

<?
error_reporting(E_ALL);
$auth = $_COOKIE["auth"];
if ($auth != "1"){
header("Location: ./signin.php");
}
$firstname = $_COOKIE['firstname'];
$id = $_COOKIE['id'];
$fname = ucwords($_COOKIE['firstname']);
$lname = ucwords($_COOKIE['lastname']);
$email = $_COOKIE['email'];
$city = ucwords($_COOKIE['city']);
$biz = ucwords($_COOKIE['biz']);

$biz_id = $_REQUEST['id'];

include("./config.php");
$result = mysql_query("SELECT * FROM company WHERE id = '$biz_id'") or die(mysql_error());
if(mysql_num_rows($result)){
    while($row = mysql_fetch_assoc($result)){
        $business_name = $row['name'];
    }
}
echo count($result);
?>

Answer 1

你没有在字符串中使用正确的变量插值，将$business_name放在qutoes中：

echo "You are claiming:$business_name as your business";

啊，看起来你也想要mysql_fetch_assoc，而不是mysql_fetch_array：

while($row = mysql_fetch_assoc($result)){
    $business_name = $row['name'];
}

Docs here

嗯...我有一个预感，试着用你的while块包裹：

if(mysql_num_rows($result)){
   ...   
}

我会让其他人在mysql_ *函数上讲课。：）

干杯

Answer 2

您需要将字符串连接起来，如下所示：

        echo "You are claiming:" . $business_name . " as your business";

Answer 3

我认为您需要在.声明中添加一些echo。

像这样：

echo "You are claiming: " . $business_name . " as your business";

Answer 4

你需要使用。连接php中的字符串

 echo "You are claiming:".$business_name ". as your business"

MySQL选择问题

4 个答案: