Question

场景是必须搜索代码，如果代码存在，结果将从MySQL DB中显示，否则显示消息“抱歉，但没有找到结果”。也许重新输入您的EVR编号：或仔细检查您的输入。“

但是，在进行搜索之前，已经显示错误消息。

我做错了什么？

<?php
    $no_results = NULL;
    $query = (isset($_POST['query']) ? $_POST['query'] : null);

    $raw_results = mysql_query("SELECT * FROM evrdata WHERE evr_no = '$query'") or die(mysql_error());

    if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following

        while($results = mysql_fetch_array($raw_results)){            
          $evr_no ='<br/>EVR No. :  '.'<b>'.$results['evr_no'].'</b>';
          $surname ='<br/>Surname :  '.'<b>'.$results['surname'].'</b>';
          $othername ='<br/>First Names :  '.'<b>'.$results['othername'].'</b>';
          $ps_code ='<br/>PS Code :  '.'<b>'.$results['ps_code'].'</b>';

        }
    }
    else { // if there is no matching rows do following
        $no_results = '<br/>Sorry, but there were no results found. Perhaps re-enter your EVR No.: or double check your entry.</b>';        
    }

?>

并在同一文件中，在HTML中显示结果：

<form action="index.php" method="POST" class="search">
       <div class="col-lg-12 col-md-12 col-sm-12 col-xs-12">
            <div class="form-group">
                <input type="text" name="query" class="form-control" placeholder="Enter Your EVR No." required/>
                 <button type="submit" id="form-submit" value="Search" class="btn-submit btn btn-big dark-blue-bordered-btn">Submit</button>
            </div>
        </div>
        <div class="col-lg-12 col-md-12 col-sm-12 col-xs-12">
            <?php 
                 if($no_results!="") // Two times Doule Quotation marks that is, and != means "not equal to". So we mean to say if $code is not equal to empty
                 {
                     echo $no_results;
                 }
                 else { echo " "; }

我尝试了isset（），empty（）但是即使没有进行搜索，错误信息仍会显示。

我错过了什么？

Answer 1

它粗糙，但那样做。原因是当加载页面时，您的搜索框为空并且正在执行查询，其中NULL返回0行。

<?php
$no_results = NULL;
$query = (isset($_POST['query']) ? $_POST['query'] : null);

if ($query) {
     $raw_results = mysql_query("SELECT * FROM evrdata WHERE evr_no = '$query'") or die(mysql_error());

     if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following

     while($results = mysql_fetch_array($raw_results)){            
      $evr_no ='<br/>EVR No. :  '.'<b>'.$results['evr_no'].'</b>';
      $surname ='<br/>Surname :  '.'<b>'.$results['surname'].'</b>';
      $othername ='<br/>First Names :  '.'<b>'.$results['othername'].'</b>';
      $ps_code ='<br/>PS Code :  '.'<b>'.$results['ps_code'].'</b>';

    }
  }
  else{ // if there is no matching rows do following
    $no_results = '<br/>Sorry, but there were no results found. Perhaps re-enter your EVR No.: or double check your entry.</b>';        
}
}

＆GT;

Answer 2

<?php
$no_results = NULL;
if (isset($_POST['query']) && strlen($_POST['query']) > 0) {
$query = (isset($_POST['query']) ? $_POST['query'] : null);

$raw_results = mysql_query("SELECT * FROM evrdata WHERE evr_no = '$query'") or die(mysql_error());

if(mysql_num_rows($raw_results) > 0){ // if one or more rows are returned do following

    while($results = mysql_fetch_array($raw_results)){            
      $evr_no ='<br/>EVR No. :  '.'<b>'.$results['evr_no'].'</b>';
      $surname ='<br/>Surname :  '.'<b>'.$results['surname'].'</b>';
      $othername ='<br/>First Names :  '.'<b>'.$results['othername'].'</b>';
      $ps_code ='<br/>PS Code :  '.'<b>'.$results['ps_code'].'</b>';

    }
}
else{ // if there is no matching rows do following
    $no_results = '<br/>Sorry, but there were no results found. Perhaps re-enter your EVR No.: or double check your entry.</b>';        
}
} else {
     //no query value
}

＆GT;

Answer 3

echo -n "somevalue" | openssl sha512 -hmac "somekey"

这应该是最好的选择...停止使用mysql_ *函数使用PDO或MySQLi。您的代码是 VULNERABLE 到sql-injection

PHP：变量显示值，即使未单击提交按钮也是如此

3 个答案: