php - Notice: Array to string conversion in [duplicate]

Question

I am trying to compose a SQL query via php. But I get the notice mentioned in a pretty unusual way.

$i=0;
$SQL="";
$tableFieldsQueryFormat=array("hospitals"=>"(location LIKE '%@val%' OR phone_no LIKE '%@val%')","doctors"=>"doc_name LIKE '%@val%'","news"=>"title LIKE '%@val%'","diseases"=>"name LIKE '%@val%'","tips"=>"tip LIKE '%@val%'");
$tableName=array("hospitals","doctors","news","diseases","tips");
$query_str=str_replace("@val",trim("dummy"),$tableFieldsQueryFormat[$tableName[$i]]);
$SQL="SELECT * FROM $tableName WHERE $query_str AND isdeleted='0';"; // Notice is thrown here !!!

I have not treated the '$SQL' variable as an array in any part of the program, in fact I only declared and used it in the above snippet. What could be the reason behind this weird notice? I don't think I've made any syntax or logic errors.

$tableName is an array and not a string, SQL queries take only string input. — Amol
– Amol, Commented Oct 25, 2016 at 22:20
@Amol told you the answer, but I'm pretty sure your making this more complicated than it needs to be. Arrays should serve a purpose. — StackSlave
– StackSlave, Commented Oct 25, 2016 at 22:24

Dave Thomas · Accepted Answer · 2016-10-25 22:22:39Z

0

You are seeing this exception because $tableName is an array. When your string is being interpolated $tableName can not be converted to a string.

Try this instead:

$SQL = "SELECT * FROM " . explode(",",$tableName) . " WHERE $query_str AND isdeleted='0';";

answered Oct 25, 2016 at 22:22

Dave Thomas

3,8452 gold badges36 silver badges44 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

php - Notice: Array to string conversion in [duplicate]

1 Answer 1

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Linked

Related