相关疑难解决方法(0)

我在这做错了什么.我已经遵循了很多例子,但似乎无法让这个工作.我有2张桌子

表=>用户

user_id
user_name
user_email
user_password
user_country
user_dobdate
user_company
user_code
user_status
user_type

表=>应用程序

apply_id
apply_from
apply_leave_type
apply_priority
apply_start_date
apply_end_date
apply_halfday
apply_contact
apply_reason
apply_status
apply_comment
apply_dated
apply_action_date

SQLI QUERY

$query = $db->select("SELECT users.user_id, app.apply_from FROM users INNER JOIN applications ON  users.user_id = app.apply_from WHERE users.user_code='1'");
$rows = $db->rows();
foreach ($rows as $apply){
$apply_id = $apply['apply_id'];
$apply_from = $apply['apply_from'];

错误信息

Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in xxxxxxxxxxxxxxx line 26

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

我在下面收到以下错误.53号线是if(mysqli_num_rows($r) == 1) {.我该如何解决这个问题.

PHP error: on line 53: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given

这是PHP MySQL代码.

$d = "UPDATE users SET del = 1, del_date = NOW() WHERE userid = '" . $_SESSION['userid'] . "'";
$r = mysqli_query ($mysqli, $d) or trigger_error("Query: $d\n<br />MySQL Error: " . mysqli_error($mysqli));

if(mysqli_num_rows($r) == 1) {

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

所以我试图检查以下查询是否返回任何结果

  $numUsersSameRatingQuery="SELECT * `user ratings` WHERE  category='$category' AND categoryId='$categoryId' AND  `userAromaRating`>0 ";

  $result=mysql_query($numUsersSameRatingQuery);

   $numResults=mysql_num_rows($result);

  if($numResults==0)
  {
   // do the INSERT INTO
  }
  else
  {
   //do the UPDATE SET
  }

但是,上面返回以下错误,并且if($numResults)每次都执行.换句话说,我永远无法else运行该块.

这是错误

 Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\boozeDb\profileThis.php on line 141

帮助表示赞赏.

可能重复:
PHP错误:mysql_fetch_array()期望参数1是资源,给定布尔值

我似乎无法弄清楚我做错了什么.因此,当我提交表单时,我收到警告错误

注意:未定义的变量:第30行/Library/WebServer/Documents/ArturoLuna_Final/loginCheck.php中的dbusername

$username = $_POST['username'];
$password = $_POST['password']; 

if($username&&$password)
{
    require 'conn.php';
    $query = "SELECT * FROM  users WHERE username='$username'";
    $result = $mysql->query($query) or die(mysqli_error($mysql));
    $numrows = $result->num_rows;

    if ($numrows!=0)
    {
        while($row = mysql_fetch_assoc($result))
        { 
            $dbusername = $row['username'];
            $dbpassword = $row['password']; 
        }

        //check to see if they match!
        if($username==$dbusername&&$password==$dbpassword)
        { 
            echo "youre In!";
        }
        else
        echo "incorrect password!";

    }
    else
        die("that user is dead");

    //echo $numrows;

}

else

    echo ("Please Enter Username")

我可能做错了什么？

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

当我使用

array_map('mysql_real_escape_string', $_POST);

it display

    Warning: mysql_real_escape_string() expects parameter 1 to be string, array given in D:\xampp\htdocs\...\...\xyz.php on line 14

之后是什么原因？

编辑: 如果我使用

array_walk_recursive($_POST, 'mysql_real_escape_string');

然后它显示

Warning: mysql_real_escape_string() expects parameter 2 to be resource, integer given in D:\xampp\htdocs\..\...\xyz.php on line 17

还请告诉我上述两种方法的区别？先感谢您

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

嗨,我正在尝试使用fgetcsv处理csv,但我所有我要去的是一个无限循环的错误

警告:第5行的testcsv.php中的fopen("Tile","User"...)

警告:fgetcsv()期望参数1是资源,在第6行的/home/ratena/public_html/proc_files/testcsv.php中给出布尔值

<?php
$uploadcsv = "/home/ratena/public_html/temp/files/BatchLoadPM15.csv";
$filecsv = file_get_contents($uploadcsv);
         //$batchid = $_POST['batchid'];
         $handle = fopen("$filecsv", "r");
         while (($data = fgetcsv($handle, 100000, ",")) !== FALSE) {
           print_r($data);   
        }
?>

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

我试图连接mysql数据库.但是我收到以下错误.

警告:: mysql_connect() [2002]无法建立连接,因为目标机器在第5行主动(尝试连接tcp://localhost:3306)test.php

警告:mysql_connect():无法建立连接,因为目标计算机主动拒绝连接.在第5行的test.php中警告:mysql_close()期望参数1是资源,test.php第15行给出布尔值

test.php的:

<?php
$link = mysql_connect(localhost, dbuser, dbpass);
if (!$link) {
   die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_close($link);
?>

所以我的代码是这个..

<?php

$password=(!isset($_POST['password']));
$username=(!isset($_POST['username']));

    $username = mysql_real_escape_string($username);
    $password = mysql_real_escape_string($password);

$query = mysql_query ("SELECT * FROM tb_funcionario WHERE username='$username' and password='$password'");

   $result = mysql_query($query);

   var_dump($result);

   $num_rows = $result->$num_rows;

    if ($num_rows)
    {
        echo "username already exist";
    }
    else
    {
     $query = "INSERT INTO tb_funcionario (nome_funcionario, username, password) VALUES (
    '$_POST[nome_funcionario]',
    '$_POST[username]',
    '$_POST[password]'
    )";;

       $result = mysql_query($query) or die (mysql_error());

    }


        mysql_query($query);
        mysql_close($bd_con);
?>

它总是给我"mysql_query()期望参数1是字符串,资源给定",我无法弄清楚如何解决它.

你能帮助我吗？

可能重复:
mysql_fetch_array()期望参数1是资源,在select中给出boolean

我正在开发一个导入MLS列表数据的PHP脚本.当我通过CLI执行脚本时,我收到以下错误:

PHP警告:curl_setopt()期望参数1是资源,字符串283中的/var/www/clients/client4/web4/web/wp-content/plugins/import-rets/RETS_class.php中给出的字符串PHP警告:curl_exec( )期望参数1是资源,字符串284中的/var/www/clients/client4/web4/web/wp-content/plugins/import-rets/RETS_class.php中给出的字符串PHP警告:curl_error()期望参数1到是资源,第287行/var/www/clients/client4/web4/web/wp-content/plugins/import-rets/RETS_class.php中给出的字符串PHP警告:curl_getinfo()期望参数1为资源,给定字符串在/var/www/clients/client4/web4/web/wp-content/plugins/import-rets/RETS_class.php第295行PHP警告:curl_close()期望参数1是资源,字符串在/ var/www中给出第191行上的/clients/client4/web4/web/wp-content/plugins/import-rets/RETS_class.php PHP警告:curl_setopt()期望参数2为long,在/ var/www/clients/client4 /中给出数组WEB4 /网络/可湿性粉剂内容/插件/进口可再生能源技术/ RETS_clas 第283行的s.php PHP警告:curl_setopt()期望参数2为long,在/var/www/clients/client4/web4/web/wp-content/plugins/import-rets/RETS_class.php中给出的数组在线283

这是我的代码:

 if (!$this->user_agent_auth )  {
 $options[CURLOPT_HTTPAUTH] = CURLAUTH_DIGEST|CURLAUTH_BASIC;
 $options[CURLOPT_USERPWD]  = $this->account.":".$this->password;
 } elseif ($this->exec_count <> 1 ) {
 $options[CURLOPT_HTTPAUTH] = CURLAUTH_DIGEST;
 $options[CURLOPT_USERPWD]  = $this->account.":".$this->password;
 }
 curl_setopt($this->ch, $options, $value);
 $response = curl_exec($this->ch);


 $error = curl_error($this->ch);
 if ( $error != '' ) {
 $this->response['curl_error'] = $error;
 $this->response['request_header'] ='';
 $this->response['received_header']='';
 $this->response['received_body']  ='';
 $this->response['http_code']      ='';
 } else {
 $this->response = curl_getinfo( $this->ch );
 $this->response['curl_error'] = '';
 $this->response['received_header'] = substr( $response, 0, …

可能重复:
PHP:警告:sort()期望参数1是数组,给定资源

我的PHP函数脚本昨晚工作正常,现在当我今天登录工作时,我得到了更多

"警告:mysql_result()期望参数1是资源,布尔值为".

我 - 不知道为什么这不起作用.我已经在线阅读了PHP手册,我甚至看到过我所做的工作和使用的例子.有人可以帮我解决这个问题吗？我一直在修复bug之后的错误(当我今天登录时很多东西都停止了工作)并且我在这里结束了我的智慧.如果有帮助,我在Windows 7上使用XAMPP作为我的服务器.

代码:( 也可通过Pastebin获得)

<?php

function dbConnect() {
$dbserver="127.0.0.1";
$dbuser="Mike";
$dbpassword="mike";
$dbname="devsite";

$con = mysql_connect($dbserver, $dbuser, $dbpassword);
mysql_select_db($dbname, $con);
}

function getSiteTitle() {


$siteTitle = mysql_result(mysql_query("SELECT \`siteTitle\` FROM siteSettings"), 0);
return $siteTitle;
}

function getSiteHeader(){

$siteHeader = mysql_result(mysql_query("SELECT \`siteHeader\` FROM siteSettings"), 0);
return $siteHeader;
}

function getBodyContent() {


$bodyContent = mysql_result(mysql_query("SELECT \`bodyContent\` FROM siteSettings"), 0);
return $bodyContent;
}

?>