在同一页面中搜索并显示结果

时间：2017-10-04 12:01:56

标签： php

我对PHP编码有一点疑问，请帮帮我。实际上我正在显示当前的作业，搜索后，结果将显示在同一页面中。已完成，但结果显示在内容下方。我只需显示该页面的结果？我想让之前的内容无法使用。这是代码：

<html>
<head>
<title>Example</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
</head>

<body>

<h2>Job Openings</h2>
<form method="POST" action="jobopenings.php">
<input type="text" name="txt"  required />
<input type="submit" name="btn" value="search" /><br/>
</form>
</body>
</html>

<?php

define('HOST', 'localhost');
define('USER', 'root');
define('PASS', '');
define('DB', '*****');


 $con = mysqli_connect(HOST,USER,PASS,DB) or die("Unable to connect to db");


$selQ = "Select * from jobpostings";
$res1 = mysqli_query($con, $selQ);

while($row = mysqli_fetch_array($res1))

echo $row[3]."<br> Job Id:<b>".$row[2]."</b><br><b>".$row[1]."</b>"."
<br>"."Exp:".$row[5]."<br>"."Location:".$row[6]."<br>".$row[8]."<br><br>";


if(isset($_POST["btn"])){

    $query=$_POST['txt'];
$query=htmlspecialchars($query);
$query=mysqli_real_escape_string($con,$query);
$raw_results="select * from jobpostings where (C_name like '%" .$query."%') 
or (Job_title like '%".$query."%')";
$final_results=mysqli_query($con,$raw_results);

if(mysqli_num_rows($final_results)>0){
    while($results=mysqli_fetch_array($final_results)){
        echo "<p><b>".$results[1]."</b><br> Job Id:".$results[2]
                                              <br>".$results[3]."</p>";
    }
  }
 else{
    echo '<b style="color:red;">No results found</b>';
    }
  }
    ?>

提前致谢

4 个答案:

答案 0 :(得分：0)

将整个php代码放在<body>中，并将if条件显示为任何部分。像这样：

<html>
<head>
<title>Example</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
</head>

<body>

<h2>Job Openings</h2>
<?php if(!isset($_POST["btn"]))
{  ?>
  <form method="POST" action="jobopenings.php">
  <input type="text" name="txt"  required />
  <input type="submit" name="btn" value="search" /><br/>
  </form>
<?php }

define('HOST', 'localhost');
define('USER', 'root');
define('PASS', '');
define('DB', '*****');


 $con = mysqli_connect(HOST,USER,PASS,DB) or die("Unable to connect to db");


$selQ = "Select * from jobpostings";
$res1 = mysqli_query($con, $selQ);

while($row = mysqli_fetch_array($res1))

echo $row[3]."<br> Job Id:<b>".$row[2]."</b><br><b>".$row[1]."</b>"."
<br>"."Exp:".$row[5]."<br>"."Location:".$row[6]."<br>".$row[8]."<br><br>";


if(isset($_POST["btn"])){

    $query=$_POST['txt'];
$query=htmlspecialchars($query);
$query=mysqli_real_escape_string($con,$query);
$raw_results="select * from jobpostings where (C_name like '%" .$query."%') 
or (Job_title like '%".$query."%')";
$final_results=mysqli_query($con,$raw_results);

if(mysqli_num_rows($final_results)>0){
    while($results=mysqli_fetch_array($final_results)){
        echo "<p><b>".$results[1]."</b><br> Job Id:".$results[2]
                                              <br>".$results[3]."</p>";
    }
  }
 else{
    echo '<b style="color:red;">No results found</b>';
    }
  }
    ?>
</body>
</html>

答案 1 :(得分：0)

以下是修改后的代码：

<?php

define('HOST', 'localhost');
define('USER', 'root');
define('PASS', '');
define('DB', '*****');


$con = mysqli_connect(HOST,USER,PASS,DB) or die("Unable to connect to db");




$final_results = null;
if(isset($_POST["btn"])){

    $query=$_POST['txt'];
    $query=htmlspecialchars($query);
    $query=mysqli_real_escape_string($con,$query);
    $raw_results="select * from jobpostings where (C_name like '%" .$query."%') or (Job_title like '%".$query."%')";
    $final_results=mysqli_query($con,$raw_results);
}
    ?>

    <html>
<head>
    <title>Example</title>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
</head>

<body>

<h2>Job Openings</h2>
<form method="POST" action="jobopenings.php">
    <input type="text" name="txt"  required />
    <input type="submit" name="btn" value="search" /><br/>
</form>

<?php

    if(!is_null($final_results) && mysqli_num_rows($final_results)>0){
        while($results=mysqli_fetch_array($final_results)){
            echo "<p><b>".$results[1]."</b><br> Job Id:".$results[2]."<br>".$results[3]."</p>";
            }
    }else{
        echo '<b style="color:red;">No results found</b>';
    }

    $selQ = "Select * from jobpostings";
    $res1 = mysqli_query($con, $selQ);

    while($row = mysqli_fetch_array($res1))

        echo $row[3]."<br> Job Id:<b>".$row[2]."</b><br><b>".$row[1]."</b>"."
    <br>"."Exp:".$row[5]."<br>"."Location:".$row[6]."<br>".$row[8]."<br><br>";
?>

</body>
</html>

我没有测试此代码，请确认它正常。

也始终在<html>标记之前启动php代码，您必须打印的所有内容必须在<body>标记内

答案 2 :(得分：0)

1）我建议你不要创建2个SQL查询（一个没有搜索，一个没有搜索）。您可以根据搜索进行一次查询并附加where子句。

2）“ ..但结果显示在内容下方。”=＆gt;将所有内容包装在<body>标记内。

更新代码 

<html>
  <head>
    <title>Example</title>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
  </head>
  <body>
    <h2>Job Openings</h2>
    <form method="POST" action="jobopenings.php">
      <input type="text" name="txt"  required />
      <input type="submit" name="btn" value="search" /><br/>
    </form>

    <?php
    define('HOST', 'localhost');
    define('USER', 'root');
    define('PASS', '');
    define('DB', '*****');

    $con = mysqli_connect(HOST, USER, PASS, DB) or die("Unable to connect to db");

    $where = "";
    $searched = false;
    if (isset($_POST["btn"]) && isset($_POST['txt'])) {
      $searched = true;
      $txt = htmlspecialchars($_POST['txt']);
      $txt = mysqli_real_escape_string($con, $txt);
      $where = "WHERE (C_name like '%" .$txt. "%') or (Job_title like '%" .$txt. "%')";
    }

    $raw_results = "SELECT * FROM `jobpostings` ".$where;
    $final_results = mysqli_query($con, $raw_results);
    if (mysqli_num_rows($final_results) > 0) {
      while ($results = mysqli_fetch_array($final_results)) {
        if(!$searched){
          echo "<p><b>".$results[1]."</b><br> Job Id:".$results[2]."<br>".$results[3]."</p>";
        } else {
          echo $results[3] . "<br> Job Id:<b>" . $results[2] . "</b><br><b>" . $results[1] . "</b>" . "<br>" . "Exp:" . $results[5] . "<br>" . "Location:" . $results[6] . "<br>" . $results[8] . "<br><br>";
        }
      }
    } else {
      echo '<b style="color:red;">No results found</b>';
    }?>
  </body>
</html>

注意：使用Prepared Statements来避免SQL Injections

答案 3 :(得分：0)

<html>
<head>
<title>Example</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
</head>

<body>
<h2>Job Openings</h2>
<form method="POST" action="jobopenings.php">
<input type="text" name="txt"  required />
<input type="submit" name="btn" value="search" /><br/>
</form>
<?php

define('HOST', 'localhost');
define('USER', 'root');
define('PASS', '');
define('DB', '*****');

$con = mysqli_connect(HOST,USER,PASS,DB) or die("Unable to connect to db"); 

if(isset($_POST["btn"])){

$query=$_POST['txt'];
$query=htmlspecialchars($query);
$query=mysqli_real_escape_string($con,$query);
$raw_results="select * from jobpostings where (C_name like '%" .$query."%') 
                                       or (Job_title like '%".$query."%')";
$final_results=mysqli_query($con,$raw_results);

if(mysqli_num_rows($final_results)>0){
while($results=mysqli_fetch_array($final_results)){
    echo "<p><b>".$results[1]."</b><br> Job Id:".$results[2]
                                          <br>".$results[3]."</p>";
   }
  }
 else{
  echo '<b style="color:red;">No results found</b>';
  }
 }
  else{
  $selQ = "Select * from jobpostings";
   $res1 = mysqli_query($con, $selQ);

  while($row = mysqli_fetch_array($res1))

  echo $row[3]."<br> Job Id:<b>".$row[2]."</b><br><b>".$row[1]."</b>"." 
 <br>"."Exp:".$row[5]."<br>"."Location:".$row[6]."<br>".$row[8]."<br><br>";}
 ?>
 </body>
 </html>
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