弹出三个错误，如何修复 [英] Three more errors pop up, how can it be fix

问题描述

hi


我已经纠正了你们向我指出的错误，但现在我还有三个错误，我不知道如何解决它。一直在网上寻找类似但无济于事的东西。这是错误。它是相同的代码。



错误no1。

注意：未定义的变量：查询在 C：\ wamp64 \\\\\\\\\\\\\\\\\\\\  70  

错误编号2：

警告：mysqli_query（）：空白查询位于 C：\ wamp64 \ www。\\\\\\\\\\\\\\\\\\\\\ -digit> 70  

错误号码3：

警告：mysqli_num_rows（）要求参数 1 为mysqli_result，布尔值为 in  C：\ wamp64 \过网站上的\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 
 
 
 
 
 
这个项目必须在明天递交，所以你知道，对我来说这是整晚。任何帮助将不胜感激。我在这里。谁能告诉我如何在我的数据库中插入和显示图像，尝试了一切，现在已经用完了想法
 
 
 
 我尝试了什么： 
 
 
 
 
 <   pre  >  <？ php 
 
 
 $ db_host = '  localhost'; 
 $ db_username = '  root'; 
 $ db_password =  ; 
 $ con = mysqli_connect（$ db_host，$ db_username，$ db_password）或 die（mysqli_connect_error（））; 
 
 mysqli_select_db（$ con，'  food'）或 die（mysqli_error（$ con））; 
 if（isset（$ _ POST ['  submit']）&& isset（ $ _GET ['  img_id']））
 {
 $ sql = < span class =code-string>  SELECT * FROM tbl_images WHERE img_id = {$ img_id}; 
 $ result = mysqli_query（$ con，$ sql）或 die（ 错误： .mysql_error（$ con））; 
 $ rowcount = mysqli_num_rows（$ result）; 
} 
 
 
 ？>  
 <   html  >  
 < ;  正文 >  
  <  表格     < span class =code-attribute> method   =  post    enctype   =  multipart / form-data   >  
 <   br /  >  
 <  输入   类型  = 文件   名称  =  image >  
 <   br /  >  <   br /  >  
 <   input     type   = 提交   名称  = 提交    value   =  upload >  
 <   / form  >  
 
 
 <？ php 
 if（isset（$ _ POST [ '  submit']）&& isset（$ _ FILES ['  file']））
 {
 if（ getimagesize（$ _ FILES ['  image'] ['  tmp_name']）== false）
 {
  echo  < span class =code-string> 请选择图像; 
} 
  else  
 {
 
 $ name = addslashes（$ _ FILES ['  image'] [' 名称]）; 
 $ image = base64_encode（file_get_contents（addslashes（$ _ FILES ['  image'] [< span class =code-string>'  tmp_name']）））; 
 saveimage（$ name，$ image）; 
} 
} 
 
函数saveimage（$ name，$ image） //  此函数将图像保存到db  
 {
  $ con  = mysqli_connect（$ db_host，$ db_username， $ db_password，  tbl_images）; 
 $ sql =   INSERT INTO tbl_images（name，image）value（'$ name，'$ image' ）; 
 $ query = mysqli_query（$ con，$ sql）; 
 
 if（$ query）
 {
  echo   成功; 
} 
 其他 
 {
  echo    not Upload; 
} 
 
} 
 displayimage（）; 
函数displayimage（） //  此函数将用于显示db中的图像 
 {
 $ db_host = '  localhost'; 
 $ db_username = '  root'; 
 $ db_password =  ; 
 
  $ con  = mysqli_connect（$ db_host，$ db_username，$ db_password）; 
 $ sql =   SELECT * FROM tbl_images; 
 $ query = mysqli_query（$ con，$ query）; 
 $ num = mysqli_num_rows（$ query）; 
 for（$ i = 0; $ i< $ num; $ i ++）
 {
 $ result = mysqli_fetch_array（$ query）; 
 $ img = $ result ['  image']; 
  echo  ' < img src = data：image; base64，'。$ img。' >'; 
} 
 mysqli_close（$ con）; 
} 
 
 ？>  
 
 <   / body  >  
  <   / html  >  
 
解决方案
 
 db_host = '  localhost'; 
 
 
 db_username = '  root'; 
 
 < blockquote> db_password =  ; 
 
 
hi


I have since corrected the error that you guys have pointed out to me but now I have three more errors and i have no idea how to fix it. been searching the net for something similiar but to no avail. here are the errors. it is the same codes.



error no1. 
Notice: Undefined variable: query in C:\wamp64\www\luana_itec244\php\dashboard.php on line 70
 



error no 2: 
Warning: mysqli_query(): Empty query in C:\wamp64\www\luana_itec244\php\dashboard.php on line 70
 



error no. 3: 
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\wamp64\www\luana_itec244\php\dashboard.php on line 71
 



this project has to be handed in tomorrow so you know, it is a whole night for me. any help would be greatly appreciated. here i what I have.  can someone tell me how do i get to insert and display images in my database, have tried everything, am now ran out of ideas



What I have tried:



<pre><?php

		
		$db_host='localhost';
		$db_username='root';
		$db_password="";
		$con=mysqli_connect($db_host,$db_username, $db_password) or die (mysqli_connect_error());
		
		mysqli_select_db($con, 'food') or die (mysqli_error($con));
		if(isset($_POST['submit']) && isset($_GET['img_id']))
		{
			$sql= "SELECT * FROM tbl_images  WHERE img_id={$img_id}";
			$result=mysqli_query($con, $sql) or die("Error:" .mysql_error($con));
			$rowcount=mysqli_num_rows($result);
		}
		
			
?>
<html>
<body>	
		<form  method="post" enctype="multipart/form-data" >
			<br/>
				<input type="file" name="image">
				<br/><br/>
				<input type="submit" name="submit" value="upload">
		</form>
			
		
<?php		
		if(isset($_POST['submit'])&& isset($_FILES['file']))
		{
			if(getimagesize($_FILES['image']['tmp_name'])== false)
			{
				echo "Please select an image";
			}
			else
			{
				
				$name=addslashes($_FILES['image']['name']);
				$image=base64_encode(file_get_contents(addslashes($_FILES['image']['tmp_name'])));				
				saveimage($name, $image);
			}			
		}
	
	function saveimage($name,$image)//this function is saving the images to the db
	{
			$con = mysqli_connect($db_host, $db_username, $db_password,"tbl_images");			
			$sql="INSERT INTO tbl_images(name,image) value('$name, '$image')";
			$query=mysqli_query($con, $sql);
			
			if($query)
			{
				echo "Success";
			}
			else
			{
				echo "not Upload";
			}
		
	}
	displayimage();	
	function displayimage()//this function will be used to display the images from the db
	{
		$db_host='localhost';
		$db_username='root';
		$db_password="";
		
		$con = mysqli_connect($db_host, $db_username, $db_password);			
		$sql="SELECT * FROM tbl_images";		
		$query=mysqli_query($con, $query);
		$num=mysqli_num_rows($query);					
		for($i=0; $i<$num; $i++)
		{
			$result=mysqli_fetch_array($query);
			$img=$result['image'];
			echo'<img src="data:image;base64, '.$img. '">';
		}
		mysqli_close($con);
	}
			
?>		
		
</body>
</html>
解决方案
db_host='localhost';

db_username='root';

db_password="";


                        
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