我不想在生成记录时在mysql中显示某些表和某些列 [英] I Don't want to show some tables and some column in mysql while generating my record

问题描述

自从我已经赋予管理员动态创建表单的权限.当他创建表单时，我要做的就是动态创建表单的表格.一切正常.现在我想显示表格以及用于生成报告的列.因为我不想显示用于生成报告的(用户登录，位置，地点)表和三列(用户ID，用户_共同ID，年) 这三列对于我动态创建的所有表都是通用的.我已经给出了我到目前为止的代码.

Since I have given to administrator to created the form dynamically.when he created form,What I have done is I have created the table for the form dynamically.Every thing is working fine.Now I want to show table and column for report generation.In that I don't want to show (userlogin,place,venue) table for report generation and three columns (user_id,user_common_id,ayear) these three column are common for all the table which I created dynamically . I have given my code what I have done so far.

指导我如何为此编写mysql查询.

Guide me How to write the mysql query for this.

<?php
    $mysqli = new mysqli("localhost", "root", "", "event");
    $result = $mysqli->query("SHOW TABLES");
    while ( $row = $result->fetch_row() )
    {
    $table = $row[0];
    echo '<h3>',$table,'</h3>';
    $result1 = $mysqli->query("SELECT * FROM $table where ayear='1'");
    if($result1) 
    {
    echo '<table cellpadding="0" cellspacing="0" class="db-table">';
    $column = $mysqli->query("SHOW COLUMNS FROM $table");
    echo '<tr>';
    while($row3 = $column->fetch_row() ) 
    {
    echo '<th>'.$row3[0].'</th>';
    }
    echo '</tr>';
    while($row2 = $result1->fetch_row() ) 
    {
    echo '<tr>';
    foreach($row2 as $key=>$value) {
    echo '<td>',$value,'</td>';
    }
    echo '</tr>';
    }
    echo '</table><br />';
    }
    }
    $mysqli->close();
    ?>

当管理员创建表单时，这是动态表和列的代码.为了更好的投影

When admin created the forms, here is code for dynamic table and columns.for better projection

$query1="create table ".$porg."(id INT NOT NULL AUTO_INCREMENT,".$VALUES.",user_id int(11),user_common_id int(11),ayear varchar(30),PRIMARY KEY (id)) ENGINE = INNODB"; 
      $result=mysql_query($query1);

推荐答案

将要跳过的表和列放入数组中，然后使用in_array()在显示名称之前对其进行测试.

Put the tables and columns you want to skip in arrays, then use in_array() to test the name before showing it.

使用fetch_assoc而不是fetch_row，以便在数组键中获得列名.

Use fetch_assoc rather than fetch_row so you get the column names in the array keys.

<?php
    $skip_tables = array('userlogin', 'place', 'venue');
    $skip_columns = array('user_id', 'user_common_id', 'ayear');
    $mysqli = new mysqli("localhost", "root", "", "event");
    $result = $mysqli->query("SHOW TABLES");
    while ( $row = $result->fetch_row() )
    {
      $table = $row[0];
      if (in_array($table, $skip_tables)) {
        continue;
      }
      echo '<h3>',$table,'</h3>';
      $result1 = $mysqli->query("SELECT * FROM $table where ayear='1'");
      if($result1) 
      {
        echo '<table cellpadding="0" cellspacing="0" class="db-table">';
        $column = $mysqli->query("SHOW COLUMNS FROM $table");
        echo '<tr>';
        while($row3 = $column->fetch_row() ) 
        {
          if (!in_array($row3[0], $skip_columns)) {
            echo '<th>'.$row3[0].'</th>';
          }
        }
        echo '</tr>';
        while($row2 = $result1->fetch_assoc() ) 
        {
          echo '<tr>';
          foreach($row2 as $key=>$value) {
            if (!in_array($key, $skip_columns)) {
              echo '<td>',$value,'</td>';
            }
          }
          echo '</tr>';
        }
        echo '</table><br />';
      }
    }
    $mysqli->close();
    ?>

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