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Don't have a clue what these mean.....

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Don't have a clue what these mean..... General SQL help. Help with MySQL, MSSQL, Oracle, mSQL, PostgreSQL, etc.

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  #1  
Old April 12th, 2004, 02:38 PM
BlackLotus BlackLotus is offline
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Don't have a clue what these mean.....
I'm getting the following error messages when I try to run my PHP code:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:program filesapache groupapachehtdocsptvisionadmininsnewpatient.php on line 47

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:program filesapache groupapachehtdocsptvisionadmininsnewpatient.php on line 48


The two lines of code that the above errors are reffering to are:

$row = mysql_fetch_array($result);
$number = mysql_num_rows($result);


Please please help....I've been beating my head against a wall for this for two days...
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  #2  
Old April 12th, 2004, 02:50 PM
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sliver sliver is offline
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RE: Don't have a clue what these mean.....
Try adding this right before those two lines:
php Code:
Original - php Code
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  #3  
Old April 12th, 2004, 02:56 PM
BlackLotus BlackLotus is offline
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RE: Don't have a clue what these mean.....
okay now it's saying this:

You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use nere ')' at line 1.... and then it repeats the above error messages...
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Old April 12th, 2004, 02:57 PM
BlackLotus BlackLotus is offline
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RE: Don't have a clue what these mean.....
but that's near impossible b/c the code on line one is simply...

<?php
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  #5  
Old April 12th, 2004, 06:58 PM
opus006 opus006 is offline
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RE: Don't have a clue what these mean.....
what does your code look like?
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Old April 12th, 2004, 08:32 PM
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RE: Don't have a clue what these mean.....
No, that was a MySQL error. That error was on line one of your query.
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  #7  
Old April 13th, 2004, 05:29 PM
BlackLotus BlackLotus is offline
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here is the code you asked about
php Code:
Original - php Code 
  1.  
  2. <?php
  3.  
  4.  
  5. include("../common/db.php");
  6. $doby="";
  7. $dobm="";
  8. $dobd="";
  9. $dobiy="";
  10. $dobim="";
  11. $dobid="";
  12. $ssn1="";
  13. $ssn2="";
  14. $ssn3="";
  15. $fname="";
  16. $lname="";
  17. $middle="";
  18. $sex="";
  19. $mrn="";
  20. $occupation="";
  21. $account_number="";
  22. $age="";
  23. $ethnicity="";
  24. $therapist="";
  25. $tot_app_visits="";
  26. $physician="";
  27. $treatid="";
  28. $diagnosis="";
  29. $handdom="";
  30. $sympextrid="";
  31. $pastmedhist="";
  32. $surgerydet="";
  33. $chiefcomplaint="";
  34. $workrelated="";
  35. $comments="";
  36. $current_medications="";
  37.  
  38.  
  39. $dob="(".$doby."-".$dobm."-".$dobd.")";
  40. $initvisitdt = "(".$dobiy."-".$dobim."-".$dobid.")";
  41. $ssn="(".$ssn1."-".$ssn2."-".$ssn3.")";
  42.  
  43. $query = "select ssn from patientmaster where ssn=$ssn";
  44. $result = mysql_query($query);
  46. $row = mysql_fetch_array($result);
  47. $number = mysql_num_rows($result);
  48. if($number == 0) {
  49.  
  50. $query2="insert into patientmaster(fname, lname, middleinit, dob, sex, ssn, MRN, Occupation, account_number, age, ethnicity, therapist, tot_app_visits) values('".$fname."','".$lname."','".$middle."','".$dob."','".$sex."','".$ssn."','".$mrn."','".$occupation."','".$account_number."','".$age."','".$ethnicity."','".$therapist."','".$tot_app_visits."')";
  51.  
  52. @$mysql_query = mysql_query($query2);
  53. if (!$mysql_query){
  54.     die(mysql_error());
  55.     }
  56.  
  57. $justinserted_id = mysql_insert_id();
  58.  
  59. if($physician=="other") $physician=$phsc_txt;
  60. if($treatid=="other") $treatid=$treat_txt;
  61. $query3="insert into patientdetails(patient_id, treatorder, physician,diagnosis, initvisit,handdominance,symptomatic, pastmedhist, surgery, chiefcomplaint,workrelated,comments, current_medications)
  62. values(".$justinserted_id.",'".$treatid."','".$physician."','".$diagnosis."', '".$initvisitdt."','".$handdom."','".$sympextrid."','".$pastmedhist."','".$surgerydet."', '".$chiefcomplaint."','".$workrelated."','".$comments."', '".$current_medications."')";
  63. $mysql_query = mysql_query($query3);
  64. if ($mysql_query){
  65.     die(mysql_error());
  66.     }
  67.  
  68. $loc1 = "assessment.php?patientid=".$justinserted_id;
  69.  
  70. echo "<script language="JavaScript">n
  71. location.href='".$loc1."';n
  72. </script>nn";
  73.  
  74.  
  75. }elseif($number != 0) {
  76. ?><br><br><br>
  77. <center>
  78. <font size="+1" face="Verdana, Arial, Helvetica, sans-serif"
  79.  
  80. color="#990000">
  81. A patient with the same SSN is already in database.
  82. </font><br><br>
  83. <input type="button" value="Back" onClick="history.back()">
  84. </center>
  85. ?>
  86. }
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  #8  
Old April 13th, 2004, 07:50 PM
-vertigo- -vertigo- is offline
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RE: Don't have a clue what these mean.....
php Code:
Original - php Code 
  1. $ssn="(".$ssn1."-".$ssn2."-".$ssn3.")";
  2.  
  3. $query = "select ssn from patientmaster where ssn=$ssn";


The query is expecting a literal at the part 'where ssn=...', but you variable $ssn is something like "(01-23-45)". It ends up like 'where ssn=(01-23-45)'. This is causing the error. Change the query to be "where ssn='$ssn'".
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