Question

我真的需要你的帮助。我试图从我的数据库中回显一个表，但它首先来自我不想要的数据库中的表名，而且css很奇怪。我希望我的表在每列中有一个不同的标题，然后数据显示在相应列标题下面的统一吗哪。

php代码在这里

    <div id="content-1">
        <div>
            <?php 

                $table = 'std_details';
                // sending query for table
                $result = mysql_query("SELECT * FROM {$table}");

                if (!$result) {

                    die ("Query to show fields from table failed");

                }

                $fields_num = mysql_num_fields($result);

                echo "<h1>All Students</h1>";
                echo "<table border='2' class='gridtable'><tr>";

                // printing table headers
                for ($i=0; $i < $fields_num; $i++) { 

                    $field = mysql_fetch_field($result);
                    echo "<td>{$field->name}&nbsp;&nbsp;</td>";

                }

                echo "</tr>\n";

                // printing table rows
                while ($row = mysql_fetch_row($result)) {

                    echo "<tr>";

                    // $row is array.... foreach( .. ) puts every element
                    // of $row to $cell variable
                    foreach ($row as $cell) {

                        echo "<td>&nbsp;\t$cell</td>";

                    }
                    echo "</tr>\n";
                }

                mysql_free_result($result)
            ?>
        </div>
    </div>

表css是

    table.gridtable {
        font-family: verdana,arial,sans-serif;
        font-size:11px;
        color:#333333;
        border-width: 1px;
        border-color: #666666;
        border-collapse: collapse;
    }
    table.gridtable th {
        border-width: 1px;
        padding: 8px;
        border-style: solid;
        border-color: #666666;
        background-color: #dedede;
    }
    table.gridtable td {
        border-width: 1px;
        padding: 8px;
        border-style: solid;
        border-color: #666666;
        background-color: #ffffff;
    }

谢谢你们。我真的需要帮助！

Answer 1

对于字段名称中的表名，您只需使用sustr($field->name, strlen ( $fieldName ........)) 对于标题，请使用<th>而非<td>，并且不要使用标签作为单元格值。试试这个

// printing table headers
            for ($i=0; $i < $fields_num; $i++) { 





                $field = mysql_fetch_field($result);

              // eliminate the tablename
              $name = substr($field->name, strlen ( $fieldName ) - strlen('yourTableName),strlen ( $fieldName ) -1  )
                // echo the header th
                echo "<th>{$name}&nbsp;&nbsp;</th>";

            }

            echo "</tr>\n";

            // printing table rows
            while ($row = mysql_fetch_row($result)) {

                echo "<tr>";

                // $row is array.... foreach( .. ) puts every element
                // of $row to $cell variable
                foreach ($row as $cell) {

                    // echo the cell td
                    echo "<td>$cell</td>";

                }
                echo "</tr>\n";
            }

Answer 2

我无法理解你为什么使用mysql而不是mysqli，这可能对你有所帮助

  <!DOCTYPE html>
    <html>
    <head>
     <style>
      table.gridtable {
        font-family: verdana,arial,sans-serif;
        font-size:11px;
        color:#333333;
        border-width: 1px;
        border-color: #666666;
        border-collapse: collapse;
       }
      table.gridtable th {
         border-width: 1px;
         padding: 8px;
         border-style: solid;
         border-color: #666666;
         background-color: #dedede;
       }
     table.gridtable td {
         border-width: 1px;
         padding: 8px;
         border-style: solid;
         border-color: #666666;
         background-color: #ffffff;
       }

    </style>
  <!--- uncomment these line if you are loading css from external file --->
  <!--- <link href="Your css file" rel="stylesheet"> --->
  </head>
<div id="content-1">
      <div>
           <?php 

            $table = 'std_details';
            // sending query for table
            $result = mysql_query("SELECT * FROM {$table}");

                if (!$result) {

                  die ("Query to show fields from table failed");

                }

               $fields_num = mysql_num_fields($result);

               echo "<h1>All Students</h1>";
               echo "<table border='2' class='gridtable'><tr>";

               // printing table headers
               for ($i=0; $i < $fields_num; $i++) { 

                $field = mysql_fetch_field($result);
                echo "<td>{$field->name}</td>";

            }

            echo "</tr>";

            // printing table rows
            while ($row = mysql_fetch_row($result)) {

                echo "<tr>";

                // $row is array.... foreach( .. ) puts every element
                // of $row to $cell variable
                foreach ($row as $cell) {

                    echo "<td>&nbsp;\t$cell</td>";

                }
                echo "</tr>";
            }
            echo "</table>";

            mysql_free_result($result)
        ?>
    </div>
</div>

PHP echo表及其css

2 个答案: