【问题标题】:Displaying a column that a user created in a form显示用户在表单中创建的列
【发布时间】:2015-04-13 11:05:09
【问题描述】:

我试图显示用户创建表单的单个列,我设置了一个表和用户会话。我需要它,以便只为创建它的用户显示该列。

这是表格

<form action="core/process.php" method="post" id="registration" >
<input type="hidden" name="formID" value="Product_Tracker" />

<input type="hidden" name="id_user" value="<?php echo $_SESSION['name_of_user']; ?>" />


<p>Name of product:<input type="text" name="Name of Product" class="input" />

<p>Please select the tests that were done on the product.</p>   
<p>In Circuit Test (ICT): Yes: <input type="radio" name="ICT" value="yes" /> No: <input type="radio" name="ICT" value="no" /></p>
<p>Visual Inspection: Yes: <input type="radio" name="Visual Inspection" value="yes" /> No: <input type="radio" name="Visual Inspection" value="no" /></p>
<p>XRAY: Yes: <input type="radio" name="XRAY" value="yes" /> No: <input type="radio" name="XRAY" value="no" /></p>
<p>Automated Optical Inspection (AOI): Yes: <input type="radio" name="AOI" value="yes" /> No: <input type="radio" name="AOI" value="no" /></p>
<!--<p>Checkbox1 <input type="checkbox" name="checkbox" value="checkbox1" /> Checkbox2: <input type="checkbox" name="checkbox" value="checkbox2" /></p>-->



<input type="submit" value="Submit" />
<p>
<a href='access-controlled.php'>Back</a>
</p>
</form>
</div>
</body>
</html>
<?php VDEnd(); ?>

我试过了,但是没用,

$con = mysql_connect("","","");
if (!$con){
die("Can not connect: " . mysql_error());
}
mysql_select_db("database",$con);
$result = mysql_query("SELECT id_user, Name_of_product FROM Product_Tracker WHERE id_user=$_SESSION['name_of_user']");
while ($row = mysql_fetch_assoc($result)) {
  echo $row['Name_of_Product'];
  echo "<br />";

做到了;

mysql_select_db("database",$con);

$id=$_SESSION['name_of_user'];  
$result = mysql_query("SELECT id_user, Name_of_product FROM Product_Tracker WHERE id_user='$id'");

while ($row = mysql_fetch_array($result))
{
     echo $row['Name_of_product'] . "<br/>";
}

mysql_query($query); 


mysql_close($con);

?>

【问题讨论】:

  • 你正在使用 mysql_* 已被贬低。使用 mysqli_*

标签: php html forms session


【解决方案1】:

你试过了吗

$id=$_SESSION['name_of_user'];    
$result = mysql_query("SELECT id_user, Name_of_product FROM Product_Tracker WHERE id_user='$id'");

【讨论】:

  • 我把它改成了mysql_select_db("database",$con); $id=$_SESSION['name_of_user']; $result = mysql_query("SELECT id_user, Name_of_product FROM Product_Tracker WHERE id_user='$id'"); while ($row = mysql_fetch_assoc($result)) { echo $row['Name_of_Product']; echo "&lt;br /&gt;"; 仍然没有显示任何东西,我认为回声有问题
  • echo "SELECT id_user, Name_of_product FROM Product_Tracker WHERE id_user='$id'";直接在mysql数据库中运行并检查您的查询是否正常
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