我怎么能修复这段代码，我很绝望 [英] How can I fix this code, I am desperate
本文介绍了我怎么能修复这段代码，我很绝望的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
尝试使用php插入和显示数据库，我有三个错误。
错误编号1.
警告：mysqli_fetch_array（）需要参数 1 是mysqli_result，boolean给出在 C：\ wamp64 \ www。\\\\\\\\\\\\\\\\\\ > 32
调用堆栈
错误编号2：
警告： move_uploaded_file（images / bhaji_n_pigtails.jpg）：无法打开流： C中没有此类文件或目录：位于 80 <\\ _wamp64 \过网站上的\\\\\\\\\\\\\\\\\\\\\\\\ br />
请帮忙，非常感谢
我尝试过：
< pre > <？ php
$ db_host = ' 本地主机';
$ 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 >
< ; body >
< div id = content >
<？ php
$ db_host = ' localhost';
$ db_username = ' root';
$ db_password = ;
$ con = mysqli_connect（$ db_host，$ db_username，$ db_password）或 die（mysqli_connect_error（））;
$ sql = SELECT * FROM tbl_images;
$ result = mysqli_query（$ con，$ sql）;
while（$ result = mysqli_fetch_array（$ result））
{
echo < div id = img_div;
echo < img src =' images /。$ row [' image']。 '>;
echo < p>。$ row [' text']。 < / p>;
echo < div>< /跨度>;
}
？>
< / div >
< 表格 < span class =code-attribute> method = post enctype = multipart / form-data >
< br / >
< 输入 type
= hidden 名称 = size value = 100000 >
< br / > < br / >
< div >
< ; 输入 typ e = file 名称 = image value = 上传 >
< / div >
< div >
< textarea name = text cols = 40 行 = 4 占位符 = 说一些 > < / textarea >
< / div >
< < span class =code-leadattribute> div >
< 输入 type = 提交 名称 = upload value = 上传 / >
< / div >
< / form >
<？ php
$ msg = ;
if（isset（$ _ POST [' upload']））
{
$ image_text = mysqli_real_escape_string（$ con，$ _ POST [' 文本']）;
$ target = images / .basename（$ _ FILES [< span class =code-string>' image'] [' 名称]）;
$ db_host = ' localhost';
$ db_username = ' root';
$ db_password = ;
$ con = mysqli_connect（$ db_host，$ db_username，$ db_password）;
$ image = $ _FILES [' image'] [' name'];
$ text = $ _POST [' text ];
$ sql = INSERT INTO tbl_images（'image'，'text'）值（$ text，$ image）;
mysqli_query（$ con，$ sql）;
if（move_uploaded_file（$ _ FILES [' image'] [' tmp_name']，$ target））
{
$ msg = 图片上传成功;
}
else
{
$ msg = 上传图片时出错;
}
$ result = mysqli_query（$ con， SELECT * FROM tbl_images）;
}
？>
< / body >
< / html >
解决方案
db_host = ' localhost';
db_username = ' root';
db_password = ;
trying to insert and display images the db using php and i have three errrors.
error no 1.
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\wamp64\www\luana_itec244\php\dashboard.php on line 32
Call Stack
error no 2:
Warning: move_uploaded_file(images/bhaji_n_pigtails.jpg): failed to open stream: No such file or directory in C:\wamp64\www\luana_itec244\php\dashboard.php on line 80
please help and much appreciated
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>
<div id="content">
<?php
$db_host='localhost';
$db_username='root';
$db_password="";
$con=mysqli_connect($db_host,$db_username, $db_password) or die (mysqli_connect_error());
$sql ="SELECT * FROM tbl_images";
$result = mysqli_query($con, $sql);
while($result = mysqli_fetch_array($result))
{
echo "<div id=img_div";
echo "<img src='images/" .$row['image']. "'>";
echo "<p>" .$row['text']. "</p>";
echo "<div>";
}
?>
</div>
<form method="post" enctype="multipart/form-data" >
<br/>
<input type="hidden" name="size" value="100000">
<br/><br/>
<div>
<input type="file" name="image" value="Upload">
</div>
<div>
<textarea name="text" cols="40" rows="4" placeholder="say something"></textarea>
</div>
<div>
<input type="submit" name="upload" value="Upload" />
</div>
</form>
<?php
$msg="";
if(isset($_POST['upload']))
{
$image_text = mysqli_real_escape_string($con,$_POST['text']);
$target="images/" .basename($_FILES['image']['name']);
$db_host='localhost';
$db_username='root';
$db_password="";
$con = mysqli_connect($db_host, $db_username, $db_password);
$image = $_FILES['image']['name'];
$text = $_POST['text'];
$sql = "INSERT INTO tbl_images ('image', 'text') values($text, $image)";
mysqli_query($con, $sql);
if(move_uploaded_file($_FILES['image']['tmp_name'], $target))
{
$msg ="Image upload successfully";
}
else
{
$msg = "Error uploading image";
}
$result = mysqli_query($con, "SELECT * FROM tbl_images");
}
?>
</body>
</html> 解决方案
db_host='localhost';
db_username='root';
db_password="";
这篇关于我怎么能修复这段代码，我很绝望的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
查看全文
