SQL，PHP：使用值以逗号分隔的另一个表输出表值

我有两个表，第一个是包含出版物的researchprojects，这些是逗号分隔的数字，另一个是publi，我有Pub_id，其中相同的值不以逗号分隔，还有Pub_year等列。我的问题是，我如何链接这两个表并输出Pub_id，Pub_year值

研究项目表

publi表

---

创建一个数组，该数组将以键和值对的形式在数组中存储发布id和发布年份

<?php 

$pub = [800 => 2015, 900 => 2016, 1000 => 2017];
foreach ($pub as $pubId => $pubYear) 
{
    echo "<td>{$pubId}, {$pubYear}</td>";
}

?>

---

以下是您的具体操作方法： 在MYSQL中链接两个表： 在MySQL中链接两个表非常容易，每个表中至少需要有一列（字段）彼此相似，才能将它们相互关联起来

例如：

<?php
// Assuming that $publications will hold all fetched values from researchprojects table
// Also assuming that $publi will hold all fetched data from publi table
$publi_ids = explode(",",$publications[0]['publications']);
//Now you have all ids in array in $publi_ids
//So moving forward with the code

foreach ($publi_ids  as $publi_id) {
echo "<h1>".$publi_id."</h1>";
$id = $publi_id;
$query = mysqli_query($mysqli,"SELECT * FROM publi WHERE publi_id='$id'");

$publi_data = array();
while ($rows = mysqli_fetch_array($query)) {
  $publi_data[] = $rows;
}

foreach ($publi_data as $publi_row) {
echo "Year :".$publi_row['publi_year']."</br>";
echo "</br></br></br>";
}
echo "</br></br></br></br>";  
} 
?>

800
Year : 2012
Year : 2013

900
Year : 2014
Year : 2015

1000
Year : 2016
Year : 2017

在您的情况下，两个表结构都需要这样：

Table : researchprojects => id,publications

Table : publi => publi_id,publi_year

您的案例：

<?php
// Assuming that $publications will hold all fetched values from researchprojects table
// Also assuming that $publi will hold all fetched data from publi table
$publi_ids = explode(",",$publications[0]['publications']);
//Now you have all ids in array in $publi_ids
//So moving forward with the code

foreach ($publi_ids  as $publi_id) {
echo "<h1>".$publi_id."</h1>";
$id = $publi_id;
$query = mysqli_query($mysqli,"SELECT * FROM publi WHERE publi_id='$id'");

$publi_data = array();
while ($rows = mysqli_fetch_array($query)) {
  $publi_data[] = $rows;
}

foreach ($publi_data as $publi_row) {
echo "Year :".$publi_row['publi_year']."</br>";
echo "</br></br></br>";
}
echo "</br></br></br></br>";  
} 
?>

800
Year : 2012
Year : 2013

900
Year : 2014
Year : 2015

1000
Year : 2016
Year : 2017

与您的情况一样，您的一行中提到的所有

id

，如下所示：

    researchprojects 
         |
         V
---------------------------
publications
800,900,1000
---------------------------

       publi 
         |
         V
---------------------------
publi_id     publi_year
800          2012
800          2013
900          2014
900          2015
1000         2016
1000         2017
---------------------------

显示表格数据输出的代码：

<?php
// Assuming that $publications will hold all fetched values from researchprojects table
// Also assuming that $publi will hold all fetched data from publi table
$publi_ids = explode(",",$publications[0]['publications']);
//Now you have all ids in array in $publi_ids
//So moving forward with the code

foreach ($publi_ids  as $publi_id) {
echo "<h1>".$publi_id."</h1>";
$id = $publi_id;
$query = mysqli_query($mysqli,"SELECT * FROM publi WHERE publi_id='$id'");

$publi_data = array();
while ($rows = mysqli_fetch_array($query)) {
  $publi_data[] = $rows;
}

foreach ($publi_data as $publi_row) {
echo "Year :".$publi_row['publi_year']."</br>";
echo "</br></br></br>";
}
echo "</br></br></br></br>";  
} 
?>

800
Year : 2012
Year : 2013

900
Year : 2014
Year : 2015

1000
Year : 2016
Year : 2017

我已经解决了这个问题

$query = "SELECT * FROM researchprojects";
$result = mysqli_query($con, $query);
while ($val = mysqli_fetch_array($result))
{
  $Id = $val['Id'];
  $Publications = $val['Publications'];
  $Publication_seprated = explode(',', $Publications);

  foreach($Publication_seprated as $Publication_seprated)
  {
  $query2 = "SELECT * FROM publi WHERE Pub_id = '$Publication_seprated'";
  $result2 = mysqli_query($con, $query2);
  while ($val2 = mysqli_fetch_array($result2))
        {
           $pub_id = $val2['pub_id'];
           $pub_year = $val2['pub_year'];

           echo $pub_year;
          } // while loop


    } // foreach

   } // while loop

---

这似乎不是MySQL表结构？：很抱歉，我找不到显示结构的方法，所以我将它放在htmlFix的第一个表中。请参阅规范化。谢谢草莓这是我想要显示的可能重复：谢谢，但是链接表然后在php中显示它的SQL查询是什么？谢谢你的回答，但是发布的id表和发布的publi_id表没有相同的值，因为发布有逗号分隔的值，所以如何我可以链接它们吗？嗯..相反，我认为您需要在publications表中每行添加一个发布，这样更好的做法是将所有发布放在一行中，这不是一个好的做法，您将在接下来的编码中遇到很多问题。。！你是对的，这会容易得多，但我不能分离这些值，因为它们与其他列相关，因此无法更改结构。不幸的是，我必须使用逗号分隔的值