使用PHP在HTML表中显示MySQL结果 [英] Use PHP to Display MySQL Results in HTML Table

问题描述

CodingBiz的更新:

我将其放入我的代码中

for($i=1;$i<=$numRows;$i++) {
    $output .= '<tr>';
    $row = $this->fetchAssoc($result);
    $colRow = $this->fetchAssoc($colResult);
    foreach($colRow as $colName) {
        $output .= "<td>".$row[$colName]."</td>";
    }
    $output .= '</tr>';
}

代替

for($i=1;$i<=$numRows;$i++) {
    $output .= '<tr>';
    $row = $this->fetchAssoc($result);
    for($j=1;$j<=$colNumRows;$j++) {
        $colRow = $this->fetchAssoc($colResult);
        $output .= "<td>".$row[$colRow["COLUMN_NAME"]]."</td>";
    }
    $output .= '</tr>';
}

这有什么问题吗?

原始帖子:

我正在PHP类中编写一个函数，以在表中显示查询结果.我自己没有构造任何表，我希望使用PHP完成所有工作.到目前为止，这是我的代码:

I'm writing a function in a PHP class to display the results of a query in a table. I'm not structuring any of the table myself, I want it everything to be done using PHP. Here is my code so far:

function allResults($table,$cols) {
    if(isset($cols)) {
        $query = "SELECT $cols FROM $table";
    }
    else {
        $query = "SELECT * FROM $table";
    }
    $result = $this->query($query);
    $numRows =  $this->numRows($result);
    $colQuery ="SELECT COLUMN_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE TABLE_SCHEMA='shareride'  AND TABLE_NAME='$table'";
    $colResult = $this->query($colQuery);
    $colNumRows = $this->numRows($colResult);

    $output = '<table class="allResults">';
    $output .= '<tr>';
    for($i=1;$i<=$colNumRows;$i++) {
        $colRow = $this->fetchAssoc($colResult);
        $output .= "<td>".$colRow["COLUMN_NAME"]."</td>";
    }
    $output .= '</tr>';
    for($i=1;$i<=$numRows;$i++) {
        $output .= '<tr>';
        $row = $this->fetchAssoc($result);
        for($j=1;$j<=$colNumRows;$j++) {
            $colRow = $this->fetchAssoc($colResult);
            $output .= "<td>".$row[$colRow["COLUMN_NAME"]]."</td>";
        }
        $output .= '</tr>';
    }
    $output .= '</table>';
    return $output;
}

如果不清楚，则query表示mysqli_query，numRows表示mysqli_num_rows，而fetchAssoc表示mysqli_fetch_assoc.数据库名称为"shareride".

In case it is unclear, query refers to mysqli_query, numRows refers to mysqli_num_rows, and fetchAssoc refers to mysqli_fetch_assoc. The database name is "shareride."

我知道我在这一行中遗漏了一些东西

I know I am missing something in this line:

$output .= "<td>".$row[$colRow["COLUMN_NAME"]]."</td>";

但是我只是不知道那是什么.现在，我正确显示了所有表列标题，并获得了正确数量的内容行，但是我无法用数据库中的实际数据填充这些行.

but I just don't know what it is. Right now, I get all the table column titles displayed correctly, and I get the correct number of content rows, but I just can't populate those rows with the actual data from the database.

我想念什么?任何帮助将非常表示赞赏！

What am I missing? Any help would be GREATLY appreciated!

推荐答案

从同一结果集中获取数据和列名

Get the data and column names from the same result set

  <?php
  $i = 0;
  $colNames = array();
  $data = array();
  while($row = ***_fetch_assoc($res)) //where $res is from the main query result not schema information
  {
     //get the column names into an array $colNames
     if($i == 0) //make sure this is done once
     {
        foreach($row as $colname => $val)
           $colNames[] = $colname;
     }

     //get the data into an array
     $data[] = $row;

     $i++;
  }

 ?>

更新:由@YourCommonSense建议替换上面的代码，并且有效，简单，较短-一种无需像我一样就不会出现循环获取列名/数组键的方法

  $data = array();
  while($row = mysql_fetch_assoc($res))
  {
     $data[] = $row;
  }

  $colNames = array_keys(reset($data))

如前所述:打印表格

 <table border="1">
 <tr>
    <?php
       //print the header
       foreach($colNames as $colName)
       {
          echo "<th>$colName</th>";
       }
    ?>
 </tr>

    <?php
       //print the rows
       foreach($data as $row)
       {
          echo "<tr>";
          foreach($colNames as $colName)
          {
             echo "<td>".$row[$colName]."</td>";
          }
          echo "</tr>";
       }
    ?>
 </table>

测试结果

您可以看到我如何将数据检索与表生成分开.它们现在相互依赖，您可以通过用静态数据填充数组来测试没有数据库的表生成情况

You can see how I separated the data retrieval from table generation. They are dependent of each other now and you can test your table generation without the database by populating the arrays with static data

您还可以将它们设置为单独的功能.

You can also make them into separate functions.

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