在网页上打印两个表格标题 [英] Two Table headings printed on webpage

问题描述

我正在做的是从两个表格中获取数据，并在下面运行脚本后打印出额外的标题然后将某些参数放入表格中。最后的echo $结果只是看看$ result中还有什么，它在额外的头部之前和头部之后打印出一个值。

  extract（shortcode_atts（array（
countryid=>''， 
），$ atts））; 
 
 
 //设置变量
 //从示例表中获取所有数据
 
 $ joburl =http：// www .x.co.za /工作/视图/; 
 $ result = mysql_query（SELECT wpjb_job.company_name，wpjb_job.job_category，wpjb_job.job_country，
 wpjb_job.job_title，wpjb_job.job_slug，
 wpjb_category.id，wpjb_category.title 
 FROM wpjb_job 
 INNER JOIN wpjb_category ON wpjb_job.job_category = wpjb_category.id 
 WHERE job_country ='$ countryid'
 AND（is_filled ='0'AND is_active ='1'）
 ORDER BY job_title）
或die（mysql_error（））; 
 
 echo< table border ='1'>; 
 echo< tr>< th>第< th>第< th>第< th< th< th>行业< / th>< / tr> 
 
 //继续获取下一行，直到没有更多的元素获得
 
为止（$ row = mysql_fetch_array（$ result））{
 
 //将每行的内容打印到表格中
 
 echo< tr>< td>; 
 echo'< a href =http://www.x.co.za/job/view/'.$row['job_slug']。'> '。$ row ['job_title']。'< / a>'; 
 echo< / td>< td>; 
 echo $ row ['company_name']; 
 echo< / td>< td>; 
 echo $ row ['title']; 
回显< / td>< / tr>; 
} 
 echo< / table>; 
 echo $ result; 
} 
  

解决方案

p>

将表头行和关闭表的标记置于以下条件中

  $ count = mysql_num_rows（$ result）; 
 
 if（$ count> 0）{
 echo< table border ='1'>; 
 echo< tr>< th>第< th>第< th>第< th< th< th>行业< / th>< / tr> 
} 
 //你的while循环
 while（...）{
 .... 
} 
 if（$ count> 0） {
 echo< / table>; 
} 
  

I have to following problem, after running script below it prints out an extra header at the end?

What I am doing is getting data from two tables, with certain parameters then placing them into a table. The echo $result at the end is just to see what is still inside $result and it prints out a value before the extra header as well as right after the header.

How can I get around that?

        extract(shortcode_atts(array(
                "countryid"=>'',
            ), $atts));


// Setting up variables
// Get all the data from the "example" table

        $joburl = "http://www.x.co.za/job/view/";
        $result = mysql_query("SELECT wpjb_job.company_name , wpjb_job.job_category , wpjb_job.job_country ,
                                        wpjb_job.job_title , wpjb_job.job_slug , 
                                        wpjb_category.id ,  wpjb_category.title 
                                    FROM wpjb_job 
                                        INNER JOIN wpjb_category ON wpjb_job.job_category = wpjb_category.id
                                        WHERE job_country='$countryid'
                                        AND(is_filled='0' AND is_active='1')
                                        ORDER BY job_title") 
                    or die(mysql_error());

        echo "<table border='1'>";
        echo "<tr> <th>Job</th> <th>Company</th> <th>Industry</th> </tr>";

// keeps getting the next row until there are no more to get

    while($row = mysql_fetch_array( $result )) {

// Print out the contents of each row into a table

        echo "<tr><td>"; 
        echo '<a href ="http://www.x.co.za/job/view/'.$row['job_slug'].'"> '.$row['job_title'].' </a>';
        echo "</td><td>"; 
        echo $row['company_name'];
        echo "</td><td>";
        echo $row['title'];
        echo "</td></tr>"; 
    } 
    echo "</table>";
    echo $result;
    }

解决方案

Ok Try as below

Put your header line and close tag of table into below condition

    $count = mysql_num_rows($result);

    if($count > 0){
      echo "<table border='1'>";
       echo "<tr> <th>Job</th> <th>Company</th> <th>Industry</th> </tr>";
    }
    //your while loop
    while(...) {
       ....
    }
    if($count > 0){
      echo "</table>";
    }  

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