PHP从数据库创建HTML表

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可能重复: 将数据库中的值输出到html表PHP

Possible Duplicate: outputting values from database into html table PHP

我正在尝试在.php文档中创建一个表，该表中填充了数据库中表中的值.但我无法使它正常工作.

I am trying to create a table in my .php document which is populated with values from a table in a database. but i cannot get it to work.

首先，它不会删除多于1行的值(在该特定日期只能有1个项目)

Firstly it is not deleting values when there are more than 1 row (There can only ever be 1 item on that particular day)

第二天，如果某天没有数据，则将其放在其之前的单元格中，这意味着该日期是错误的一天.

Secondly if there is no data for a particular day it just puts it into the cell before it, meaning it is on the wrong day.

这是代码:

<?php if(!empty($_POST['recipe'])) { $week = mysql_real_escape_string($_POST['week']); $day = mysql_real_escape_string($_POST['day']); $mealtime = mysql_real_escape_string($_POST['mealtime']); $recipe = mysql_real_escape_string($_POST['recipe']); $check = mysql_query("SELECT * FROM menu WHERE dayid = '".$day."' AND mealtimeid = '".$mealtime."'"); if(mysql_num_rows($check) == 1) { mysql_query("DELETE FROM menu WHERE mealtimeid = '".$mealtime."' AND dayid = '".$day."'"); $success = mysql_query("INSERT INTO menu (weekid, dayid, mealtimeid, recipeid) VALUES('".$week."', '".$day."', '".$mealtime."', '".$recipe."')"); if($success) { echo "<h1>Success</h1>"; echo "<p>Your recipe was successfully added.</p>"; } else { echo "<h1>Error</h1>"; echo "<p>Sorry there was a problem, please try again.</p>"; } } else { $success = mysql_query("INSERT INTO menu (weekid, dayid, mealtimeid, recipeid) VALUES('".$week."', '".$day."', '".$mealtime."', '".$recipe."')"); if($success) { echo "<h1>Success</h1>"; echo "<p>Your recipe was successfully added.</p>"; } else { echo "<h1>Error</h1>"; echo "<p>Sorry there was a problem, please try again.</p>"; } } } if(!empty($_POST['selectweek'])) { $selectweek = mysql_real_escape_string($_POST['selectweek']); function ouptutMeal($selectweek, $mealtime, $mealname) { $sqlmeasurement2 = mysql_query("SELECT title, dayid FROM recipe JOIN menu ON recipe.recipeid = menu.recipeid WHERE menu.weekid = '$selectweek' AND menu.mealtimeid = '$mealtime' ORDER BY dayid"); echo "<br/> <table> <td></td> <td><strong>Monday</strong></td> <td><strong>Tuesday</strong></td> <td><strong>Wednesday</strong></td> <td><strong>Thursday</strong></td> <td><strong>Friday</strong></td> <td><strong>Saturday</strong></td> <td><strong>Sunday</strong></td> <tr> <td><strong>$mealname</strong></td>"; while($info2 = mysql_fetch_array( $sqlmeasurement2 )) { if(empty($info2['dayid'])) { echo '<td></td>'; } elseif($info2['dayid'] == '1') { echo ' <td>', $info2['title'], '</td>'; } elseif($info2['dayid'] == '2') { echo ' <td>', $info2['title'], '</td>'; } elseif($info2['dayid'] == '3') { echo ' <td>', $info2['title'], '</td>'; } elseif($info2['dayid'] == '4') { echo ' <td>', $info2['title'], '</td>'; } elseif($info2['dayid'] == '5') { echo ' <td>', $info2['title'], '</td>'; } elseif($info2['dayid'] == '6') { echo ' <td>', $info2['title'], '</td>'; } else { echo ' <td>', $info2['title'], '</td>'; } } echo '</tr> </table>'; } ouptutMeal($selectweek, 1, 'Breakfast'); ouptutMeal($selectweek, 2, 'Lunch'); ouptutMeal($selectweek, 3, 'Evening Meal'); ouptutMeal($selectweek, 4, 'Pudding'); ouptutMeal($selectweek, 5, 'Supper & Snacks'); } } else { ?>

这是它从以下位置获取数据的形式:

This is the form it gets the data from:

<form method="post" action=""> <fieldset> <label for="week">Select Week:</label> <select name="week"> <option value="0"> Select Week<?php echo $item; ?> </option> </select> <label for="day">Select Day:</label> <select name= "day"> <option value="0"> Select Day<?php echo $item2; ?> </option> </select><br /> <br /> <label for="mealtime">Select Meal Time:</label> <select name= "mealtime"> <option value="0"> Select Meal Time<?php echo $item3; ?> </option> </select><br /> <br /> <label for="recipe">Select Recipe:</label> <select name="recipe"> <option value="0"> Select Recipe<?php echo $item4; ?> </option> </select> <input type="submit" id="login-submit" value="Add to Menu" /> </fieldset> </form> <form method="post" action=""> <label for="selectweek">Select Week:</label> <select name= "selectweek"> <option value="0"> Select Week<?php echo $item; ?> </option> </select> <input type="submit" id="login-submit" value="View Menu" /> </form>

-末尾的项目是在星期日，但由于前一天而落后没有项目.我将如何使该商品进入星期日，同时在其他商品与其他商品之间保持间隔.

-- The item on the end is meant to be on Sunday but is behind because a previous day does not have a item. How would i make that item go to Sunday, while keeping a gap where the other item isn't.

推荐答案

使用是否检查值是否为空.像这样的东西.

Use if to check if value is empty or not. Some thing like this.

if($info2['dayid'] == "") { echo '<td>&nbsp;</td>'; }

它将正确填充它，而不会干扰其布局.

it will fill it properly, without disturbing it layout.

希望这会有所帮助.