redarrow Posted November 14, 2010 Share Posted November 14, 2010 Advance thank you. Can you help please. The error..... Warning: mysql_fetch_assoc() expects parameter 1 to be resource, string given in C:\wamp\www\test_dabase.php on line 24 code. <?php //database connection. $DB = mysql_connect("localhost","root") or die(mysql_error()); if($DB){ //database name. $DB_NAME="mysql"; //select database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name"); // if connection. }if($CON){ //show tables. $mysql_show="SHOW TABLES"; //select show and show. $mysql_select2="mysql_query(".$mysql_show.") or die(mysql_error())"; } //if allowed to show. if($mysql_select2){ //while it and while($data=mysql_fetch_assoc($mysql_select2)){ //show it. echo $data; } } ?> Link to comment https://forums.phpfreaks.com/topic/218618-code-help-view-tables-form-a-database-via-php-code-included/ Share on other sites More sharing options...
redarrow Posted November 14, 2010 Author Share Posted November 14, 2010 Had to be done this way, with a array. <?php $DB = mysql_connect("localhost","root") or die(mysql_error()); //database name. $DB_NAME="mysql"; //selEct database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease chage database name"); $showtablequery = "SHOW TABLES FROM mysql"; $showtablequery_result = mysql_query($showtablequery); while($showtablerow = mysql_fetch_array($showtablequery_result)) { echo $showtablerow[0]."<br />"; } ?> Link to comment https://forums.phpfreaks.com/topic/218618-code-help-view-tables-form-a-database-via-php-code-included/#findComment-1134000 Share on other sites More sharing options...
redarrow Posted November 14, 2010 Author Share Posted November 14, 2010 Code corrected and understood why.... from the previous attempt, I didn't use a array, i used a assoc, so it went wrong, also I wrote the select statement wrong. also i didn't select the database name (from mysql) working <?php //database connection. $DB = mysql_connect("localhost","root") or die(mysql_error()); if($DB){ //database name. $DB_NAME="mysql"; //select database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name"); // if connection. }if($CON){ //show tables. $mysql_show="SHOW TABLES FROM $DB_NAME"; //select show and show. $mysql_select2=mysql_query($mysql_show) or die(mysql_error()); } //if allowed to show. if($mysql_select2){ //while it and while($data=mysql_fetch_array($mysql_select2)){ //show it. echo " $data[0] <BR>"; } } ?> not working.(with my mistakes)))) <?php //database connection. $DB = mysql_connect("localhost","root") or die(mysql_error()); if($DB){ //database name. $DB_NAME="mysql"; //select database and name. $CON=mysql_select_db($DB_NAME,$DB)or die(mysql_error()."\nPlease change database name"); // if connection. }if($CON){ //show tables. $mysql_show="SHOW TABLES"; //select show and show. $mysql_select2="mysql_query(".$mysql_show.") or die(mysql_error())"; } //if allowed to show. if($mysql_select2){ //while it and while($data=mysql_fetch_assoc($mysql_select2)){ //show it. echo $data; } } ?> Link to comment https://forums.phpfreaks.com/topic/218618-code-help-view-tables-form-a-database-via-php-code-included/#findComment-1134004 Share on other sites More sharing options...
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