Question

<?php 

if ($_SERVER["REQUEST_METHOD"] == "POST") {

$firstname=$lastname = $email = $address = $hphone = $mphone = "";
$link = mysql_connect('shunurcom.fatcowmysql.com', 'sachin', 'sacjap'); 

 mysql_select_db(cycle,$link);
if (!empty($_POST["firstname"])) 
{
$firstname = $_POST["firstname"];
$sql= "SELECT * FROM users WHERE firstname = '$firstname' ";
$comments=mysql_query($sql,$link);
if(!$comments==false )
{


 while($row = mysql_fetch_array($comments, MYSQL_ASSOC))
{
$firstname = $row['firstname'];
$lastname = $row['lastname'];
 $email = $row['email'];
$address = $row['address'];
$hnum = $row['hphone'];
$mnum = $row['mphone'];


 echo "  <div style='margin:30px 0px;'>
  First Name: $firstname<br />
 Last Name: $lastname<br />
  Email: $email<br />
  Address: $address<br />
  Home Number: $hnum<br />
  Mobile Number: $mnum
   </div>
  ";
}

}
else {
echo "invalid";
}
}

我尝试使用size of，mysql_num_rows但代码不会转到其他部分，当给出错误的输入时，该部分应该打印无效。

知道为什么吗？当输入无效时，它只显示一个空白屏幕。

Answer 1

我在你的代码中看到很多混淆。

请试试这个：

<?php

if($_POST)
{

    $link = mysql_connect('shunurcom.fatcowmysql.com', 'sachin', 'sacjap'); 
    mysql_select_db(cycle,$link);

    if (isset($_POST["firstname"])) 
    {
        $firstname = $_POST["firstname"];
        $sql= "SELECT * FROM users WHERE firstname = '$firstname' ";
        $comments=mysql_query($sql,$link);

       if($comments ) {
        $number=mysql_num_rows($comments);

      while($row = mysql_fetch_assoc($comments)) 
      {
        $firstname = $row['firstname'];
        $lastname = $row['lastname'];
        $email = $row['email'];
        $address = $row['address'];
        $hnum = $row['hphone'];
        $mnum = $row['mphone'];

        echo "  <div style='margin:30px 0px;'>
        First Name: $firstname<br />
        Last Name: $lastname<br />
        Email: $email<br />
        Address: $address<br />
        Home Number: $hnum<br />
        Mobile Number: $mnum
         </div> ";
       }//end of while loop.

  }
?>

Answer 2

首先mysql_query将在错误时返回FALSE，而不是在结果集为空时，第二，您应该停止使用mysql_* api，因为它已被弃用。

此扩展在PHP 5.5.0中已弃用，并已在PHP 7.0.0中删除

使用mysqli_*或更好PDO

像这样更新你的代码：

// test here if you have a result set or not
$rows = mysql_num_rows($comments);
if ($rows == 0) {
    print "invalid or no results"; // this way you won't execute the 
                                   // while block on an empty result set
}
else {
    while($row = mysql_fetch_array($comments, MYSQL_ASSOC)) {
        $firstname = $row['firstname'];
        $lastname = $row['lastname'];
        $email = $row['email'];
        $address = $row['address'];
        $hnum = $row['hphone'];
        $mnum = $row['mphone'];

        echo "<div style='margin:30px 0px;'>
            First Name: $firstname<br />
            Last Name: $lastname<br />
            Email: $email<br />
            Address: $address<br />
            Home Number: $hnum<br />
            Mobile Number: $mnum
            </div>";
    }
}

Answer 3

<?php

if ($_SERVER["REQUEST_METHOD"] == "POST")
{

    $firstname = $lastname  = $email     = $address   = $hphone    = $mphone    = "";
    $link      = mysql_connect('shunurcom.fatcowmysql.com', 'sachin', 'sacjap');

    mysql_select_db(cycle, $link);
    if (!empty($_POST["firstname"]))
    {
        $firstname = $_POST["firstname"];
        $sql       = "SELECT * FROM users WHERE firstname = '$firstname' ";
        $comments  = mysql_query($sql, $link);
        if (!$comments == false)
        {

            $number = mysql_num_rows($comments);
            if ($number >= 1)
            {
                while ($row = mysql_fetch_array($comments, MYSQL_ASSOC))
                {
                    $firstname = $row['firstname'];
                    $lastname  = $row['lastname'];
                    $email     = $row['email'];
                    $address   = $row['address'];
                    $hnum      = $row['hphone'];
                    $mnum      = $row['mphone'];


                    echo "  <div style='margin:30px 0px;'>
                        First Name: $firstname<br />
                       Last Name: $lastname<br />
                        Email: $email<br />
                        Address: $address<br />
                        Home Number: $hnum<br />
                        Mobile Number: $mnum
                         </div>
                        ";
                }
            }
            else
            {
                echo "No record Found";
            }
        }
        else
        {
            echo "invalid";
        }
    }

无法检查查询是否返回一行

3 个答案: