2017-10-05 27 views
0

我不知道正确的语法从数据库中打印出来的值获取值按钮没有得到提供错误的参数无效后提交..请帮助如何打印PHP语法HTML

<form action="index.php" method="post"> 
<input type="text" name="user_name" placeholder="User Name"> 
<input type="text" name="first_name" placeholder="First Name"> 
<input type="email" name="email" placeholder="email"> 
<input type="submit" name='submit' value="Submit"> 
<input type="submit" name='seeDatabase' value="Get Values"> 
</form> 

<table> 
<tr> 
    <th>User_Name </th> 
    <th>First_Name </th> 
    <th>Email</th> 
</tr> 
<?php 
$sqlSelect = "Select * From user"; 
$myResult2 = $connection->query($sqlSelect); 
foreach ($myResult2 as $item): ?> 
<tr> 
    <td><?php echo $item['user_name'] ?></td> 
    <td><?php echo $item['first_name'] ?></td> 
    <td><?php echo $item['email'] ?></td> 
    <td> 
     <form action="delete.php" method="post"> 
     <input type="hidden" name="userName" value="<?php echo      
     $item['user_name'] ?>"> 
     <input type="submit" name="delete" value="Delete"> 
     </form> 
    </td> 
</tr> 
<?php endforeach; ?> 
</table> 
+0

你到底想要做什么?你alredy打印数据库值内HTML'值=“<?php echo $ item ['user_name']?>”' –

+0

我想用isset()函数进行数据检查,并没有任何信息打印,直到按钮点击“获取值”的值。我通常会这样做,如果声明并将所有需要的代码放在该语句中,但是当我这样做时,它会导致html代码出现问题。这是一个im试图完成的例子 – Jason

+0

<?php if(isset($ _ POST ['seeDatabase'])) { $ sqlSelect =“Select * From user”; $ myResult2 = $ connection-> query($ sqlSelect); 的foreach($ myResult2为$项):> ​​ ​​ ​​<????? php echo $ item ['email']?> ​​ – Jason

回答

1

你在循环之前需要检查数组是否为空以防止获得无效的参数错误

if(!empty($myResult2) && count($myResult2) > 0) { 
//you can loop over 
foreach ($myResult2 as $item) { 
//foreach stuff 
} 
} else { 
//array is empty 
echo 'no result found'; 
}