在mysql_fetch_array.mysql_fetch_array（）中获取错误需要参数1为资源 [英] Getting error in mysql_fetch_array.mysql_fetch_array() expects parameter 1 to be resource

问题描述

以下是我的代码。我收到以下错误



警告：mysql_fetch_array（）要求参数1为资源，布尔值在C：\Program Files（x86）\ Ampmpth中给出第174行\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ （$ getselect））



我的尝试：



< ;？php

if（isset（$ _ GET ['ID']））

{

$ ID = $ _ GET ['ID' ];

$ getselect = mysql_query（SELECT * FROM booking_p WHERE ID ='$ ID'）;

while（$ profile = mysql_fetch_array（$ getselect））

{

$ firstname = $ profile ['firstname'];

$ email = $ profile ['email'];

$ phone = $ profile ['phone'];

$ city = $ profile ['address1'];

$ date = $ profile ['date '];

$ time = $ profile ['time'];

$ amount = $ profile ['amount'];

$ productinfo = $ profile ['productinfo'];

}}？>

解决方案

profile = mysql_fetch_array（

getselect））



我尝试过：



<？php

if（isset（

_GET ['ID']））

{

The following is my code. i'm getting following error

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\Program Files (x86)\Ampps\www\pandit\test_editform.php on line 174

the problem is exactly n this line
while($profile=mysql_fetch_array($getselect))

What I have tried:

<?php
if(isset($_GET['ID']))
{
$ID=$_GET['ID'];
$getselect=mysql_query("SELECT * FROM booking_p WHERE ID='$ID'");
while($profile=mysql_fetch_array($getselect))
{
$firstname=$profile['firstname'];
$email=$profile['email'];
$phone=$profile['phone'];
$city=$profile['address1'];
$date=$profile['date'];
$time=$profile['time'];
$amount=$profile['amount'];
$productinfo=$profile['productinfo'];
} } ?>

解决方案

profile=mysql_fetch_array(

getselect))

What I have tried:

<?php
if(isset(

_GET['ID']))
{

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