For-loop and variables in PHP MySQL Query

Question

I have to select a lot of tables and Fields using PHP and MySQL. But the structure and naming of these are every time the same.

I have the tables field_revision_blabla and the fields blablabla_value from table-alias blablabla_table.

for ($i=0; $i < sizeof($typeFields); $i++) {
    $query->join(
        "field_revision_".$typeFields[$i], 
        $typeFields[$i]."_table", 
        'n.nid = $typeFields[$i]."_table.entity_id"');
    $query->addField($typeFields[$i]."_table", $typeFields[$i]."_value");
}

But that gives me the error:

SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '[$i])."_table.entity_id" INNER JOIN field_revision_field_text field_text_table O' at line 2

I think the error is in

'n.nid = $typeFields[$i]."_table.entity_id"'

Aaron W. · Accepted Answer · 2013-07-02 11:41:14Z

2

SO's color coding should have flagged this for you...along with the SQL statement complaining about variables in the query. Your $typeFields[$i] isn't being figured in the string.

"n.nid = " . $typeFields[$i] . "_table.entity_id"

answered Jul 2, 2013 at 11:41

Aaron W.

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mirkobrankovic · Accepted Answer · 2013-07-02 11:41:22Z

1

Use double quotes instead of single, I doubt you have record in DB containing n.nid = $typeFields[$i]."_table.entity_id":

for ($i=0; $i < sizeof($typeFields); $i++) {
    $query->join(
        "field_revision_".$typeFields[$i], 
        $typeFields[$i]."_table", 
        "n.nid = ".$typeFields[$i]."_table.entity_id");
    $query->addField($typeFields[$i]."_table", $typeFields[$i]."_value");
}

answered Jul 2, 2013 at 11:41

mirkobrankovic

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