i believe code wrong selected fields aren't inserting table/fields @ all.
how following work following refernces url.
$sql="insert instruction (physioreference, physio, physiosaddress, postcode, physiomobile, number, physiofax, physiosemail reference='$reference') select physioreference, name, line1, postcode, mobile, tel, fax, email `physio` physioreference='$physioreference'";
example url: http://test.com/physiotoinstruction.php?physioreference=100099&reference=456789
page.php
<?php require_once('auth.php'); $host=""; // host name $username=""; // mysql username $password=""; // mysql password $db_name=""; // database name $tbl_name="instruction"; // table name // connect server , select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select db"); $reference=mysql_real_escape_string($_get['reference']); $physioreference=mysql_real_escape_string($_get['physioreference']); $sql="insert instruction (physioreference, physio, physiosaddress, postcode, physiomobile, number, physiofax, physiosemail reference='$reference') select physioreference, name, line1, postcode, mobile, tel, fax, email `physio` physioreference='$physioreference'"; $result=mysql_query($sql); //$sql="insert triage (reference, forename) //select reference, forename `instruction` //where reference='$reference' limit 1"; echo "successful"; echo "<br>"; echo "<a href='insert.php'>view result</a>"; // mysql_error() ?>
mysql insert
syntax not support where
clause query stands fail.
try substitute physiosemail reference='$reference')
physiosemail)
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