Php 网站上的搜索功能-不显示任何数据_Php_Html_Sql_Search

Php 网站上的搜索功能-不显示任何数据

php html sql search

Php 网站上的搜索功能-不显示任何数据,php,html,sql,search,Php,Html,Sql,Search,我已经创建了一个搜索页面，但无论我如何更改它，我都无法让它返回结果，即使它们与数据库完全匹配。也许有人能成为另一双眼睛，为我指出正确的方向吗在SQL语句中放入变量时，我总是很不确定，所以可能是错误 Searchform.php <?php include 'index.php'; ?> <html> <center> <link rel="stylesheet" type="text/css" href="skeleton.css" /> <

我已经创建了一个搜索页面，但无论我如何更改它，我都无法让它返回结果，即使它们与数据库完全匹配。也许有人能成为另一双眼睛，为我指出正确的方向吗

在SQL语句中放入变量时，我总是很不确定，所以可能是错误

Searchform.php

<?php
include 'index.php';
?>
<html>
<center>
<link rel="stylesheet" type="text/css" href="skeleton.css" />
<link rel="stylesheet" type="text/css" href="normalize.css" />
 <title>Search</title> 

<h3>Search People</h3> 
<p>You  may search either by first or last name</p> 
<form  method="post" action="search.php"  id="searchform"> 
<input  type="text" name="name"> 
<input  type="submit" name="submit" value="Search"> 
</form>       
</html> 
</p>
</center>

<?php 
    include 'pagetemplate.php';
    if(isset($_POST['submit'])){ 
        if(isset($_GET['go'])){ 
            if(preg_match("/[A-Z  | a-z]+/", $_POST['name'])){ 
                $name=$_POST['name']; 
                //-query  the database table 

              $result = $mysqli_conn->query("SELECT * FROM users WHERE FirstName LIKE '%$name%' OR LastName LIKE '%$name%'");
             print " Query returned ". $result->num_rows . " rows ";
             //The function to display data is below
             displayData($result);


                // $stmt = mysqli_prepare($con, 'SELECT * FROM users WHERE (FirstName LIKE ? OR LastName LIKE ? OR Username LIKE ?)';

                //mysqli_stmt_bind_param($stmt, 'sss', "%$name%", "%$name%", "%$name%");

                //mysqli_stmt_execute($stmt);

                //-run  the query against the mysql query function 


                } 
                else{ 
                    echo  "<p>Please enter a search query</p>"; 
                    } 
            } 
        } 
    ?>


搜寻
搜身
您可以按名字或姓氏搜索

Search.php

<?php
include 'index.php';
?>
<html>
<center>
<link rel="stylesheet" type="text/css" href="skeleton.css" />
<link rel="stylesheet" type="text/css" href="normalize.css" />
 <title>Search</title> 

<h3>Search People</h3> 
<p>You  may search either by first or last name</p> 
<form  method="post" action="search.php"  id="searchform"> 
<input  type="text" name="name"> 
<input  type="submit" name="submit" value="Search"> 
</form>       
</html> 
</p>
</center>

<?php 
    include 'pagetemplate.php';
    if(isset($_POST['submit'])){ 
        if(isset($_GET['go'])){ 
            if(preg_match("/[A-Z  | a-z]+/", $_POST['name'])){ 
                $name=$_POST['name']; 
                //-query  the database table 

              $result = $mysqli_conn->query("SELECT * FROM users WHERE FirstName LIKE '%$name%' OR LastName LIKE '%$name%'");
             print " Query returned ". $result->num_rows . " rows ";
             //The function to display data is below
             displayData($result);


                // $stmt = mysqli_prepare($con, 'SELECT * FROM users WHERE (FirstName LIKE ? OR LastName LIKE ? OR Username LIKE ?)';

                //mysqli_stmt_bind_param($stmt, 'sss', "%$name%", "%$name%", "%$name%");

                //mysqli_stmt_execute($stmt);

                //-run  the query against the mysql query function 


                } 
                else{ 
                    echo  "<p>Please enter a search query</p>"; 
                    } 
            } 
        } 
    ?>

能否尝试更改：-$result=$mysqli\u conn->query（“从名为“%$name%”或姓为“%$name%”的用户中选择*”这个到这个<代码>$result=$mysqli_conn->query（'SELECT*来自FirstName（如“%$name%”）或LastName（如“%$name%”）的用户）仍然显示空白页！我以“abc”为例，将一些数据和所有abc一起输入数据库。我想不出来！如果我使用我在下面写的注释掉的OO代码，我将如何显示它？这是我没有将那部分代码用作未完成代码的唯一原因。你能给出搜索表单的源代码吗？你确定脚本肯定会被调用吗？它的执行路径是什么。我把echo
s放得到处都是。