This question already has an answer here:
这个问题在这里已有答案:
- mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource 31 answers
mysql_fetch_array()/ mysql_fetch_assoc()/ mysql_fetch_row()/ mysql_num_rows等...期望参数1为资源31答案
$result = mysql_query("SELECT name FROM internet_shop");
while($row=mysql_fetch_assoc($result))
Its shows errors mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in
它显示错误mysql_fetch_assoc()期望参数1是资源,布尔值在
Databsename:piashhas_piash
One table in database: internet_shop
数据库中的一个表:internet_shop
Runs well in local host Xampp but when on server it shows error can anyone help thanks
在本地主机Xampp中运行良好,但在服务器上显示错误,任何人都可以帮助谢谢
1 个解决方案
#1
1
Alter the query line to
更改查询行
$result = mysql_query("SELECT name FROM internet_shop") or exit(mysql_error());
Is an error displayed? As ccKep stated, MySQL is deprecated; consider MySQLi.
是否显示错误?正如ccKep所说,MySQL已被弃用;考虑MySQLi。
$connectionId = mysqli_connect($host,$user,$pass) or exit(mysqli_connect_error());
mysqli_select_db($connectionId, "piashhas_piash") or exit(mysqli_error($connectionId));
$result = mysqli_query($connectionId, "SELECT `name` FROM `internet_shop`");
while($row=mysqli_fetch_assoc($result))
#1
1
Alter the query line to
更改查询行
$result = mysql_query("SELECT name FROM internet_shop") or exit(mysql_error());
Is an error displayed? As ccKep stated, MySQL is deprecated; consider MySQLi.
是否显示错误?正如ccKep所说,MySQL已被弃用;考虑MySQLi。
$connectionId = mysqli_connect($host,$user,$pass) or exit(mysqli_connect_error());
mysqli_select_db($connectionId, "piashhas_piash") or exit(mysqli_error($connectionId));
$result = mysqli_query($connectionId, "SELECT `name` FROM `internet_shop`");
while($row=mysqli_fetch_assoc($result))