$result=$DB_site->query("SHOW tables");
while(list($key,$val)=each($tables)){
$gettable .="'$key',";
}
$gettable = substr($gettable, 0, -1);
$alls = array($gettable);
while(list($key,$val)=each($tables)){
$gettable .="'$key',";
}
$gettable = substr($gettable, 0, -1);
$alls = array($gettable);
这种定义方法可以吗?如果不行的话试试这样:
$result=$DB_site->query("SHOW tables");
while(list($key,$val)=each($tables)){
$gettable .="'$key',";
}
$gettable = substr($gettable, 0, -1);
$alls = explode(',' $gettable);
$arr=array();
while(list($key,$val)=each($tables)){
$arr[]="'$key'";
}print_r($arr);
while(list($key,$val)=each($tables)){
$gettable .="'$key',";
}
$gettable = substr($gettable, 0, -1);
$alls = explode(',' $gettable);一写出去,就报语法错误
$result= $DB_site->query("SHOW CREATE TABLE $table");
while($listtable = $DB_site->fetch_array($result)){
$tabledump .= $listtable[1].";\n\n";
}如果用
$alls = array('news','newclass','newsuser','views');
就没有问题
但是如果使用
$result=$DB_site->query("SHOW tables");
while(list($key,$val)=each($tables)){
$gettable .="'$key',";
}
$gettable = substr($gettable, 0, -1);
$alls = array($gettable);的话,就会报错,错误提示如下:
Invalid SQL: SHOW CREATE TABLE 'news','newclass','newsuser','views'mysql 错误: You have an error in your SQL syntax near ''news','newclass','newsuser','views'' at line 1
mysql 错误号: 1064
我找不到解决的办法,可能就如“xizi2002(☆☆戏☆子☆☆) ”所说的一样
“对了$alls = array($gettable);
这种定义方法可以吗?
" -----这样定义应该不行
请各位帮忙,谢谢
则:
$gettable = substr($gettable,0,-1);
eval("\$alls = array($gettable);");
将产生数组$alls :
Array
(
[0] => news
[1] => newclass
[2] => newsuser
[3] => views
)
$alls = explode(',', $gettable);
请教: xizi2002(☆☆戏☆子☆☆) 我用你的方法,还是不行,报错如下:
Invalid SQL: SHOW CREATE TABLE 'news'
mysql 错误: You have an error in your SQL syntax near ''access'' at line 1mysql 错误号: 1064谢谢戏子继续给予帮助
<?php
$link1=@mysql_connect("192.168.0.1","dbname","******");
$dbname="dbname";
$jh = mysql_select_db($dbname);
//检测是否激活是否成功
$result = mysql_query("SHOW TABLES FROM $dbname");
while ($row=mysql_fetch_array($result)) {
print "<pre>";
print_r($row);
print "</pre>";
}