mysql - php pass value to multiple select queries -


i'm having trouble passing variable between mysql queries on same page. maybe can advise i'm doing wrong. i'm new php/mysql, answer seems easy, don't see it. here have:

1. mysql: table a:

    id | gene_id | protein_id | disease_id | etc     ----------------------------------------------     1  | 672     | p12803     | 091312     2  | 817     | p99613     | 020346     3  | 411     | p52021     | 055823

2. search result page. displays list of results. reaults identified [$id] , <a href> link passes [$id] page result details. works perfectly.

3. details page. query result search results page , display related information table, identified [$id]. works fine.

    <?php      $sql = "select * table_a     id=" . $_get["id"];      $rs_result = mysql_query ($sql,$connect);       while ($row = mysql_fetch_assoc($rs_result)) {      ?>      <table class="table">     <tr><td>gene: </td><td><? echo $row[gene_id]; ?></td></tr>     <tr><td>protein: </td><td><? echo $row[protein_id]; ?></td></tr>     <tr><td>disease: </td><td><? echo $row[disease_id]; ?></td></tr>     </table>      <?  } ?>

4. show related data. on details page, want show related data table b in same database, using "protein_id" query above. can't work, pass "protein_id" next query, follows:

table b:

    id | protein_asc | synonym | name | etc     ----------------------------------------------     11  | p12803    | abc      | |     12  | p99613    | def      | |      <?php      $new_id = $row[protein_id];      $sqla = "select * table_b     protein_asc='".$new_id."'";      $rsa_result = mysql_query ($sqla,$connect);       while ($row = mysql_fetch_assoc($rsa_result)) {      ?>      <table class="table">     <tr><td>synonym: </td><td><? echo $row[synonym]; ?></td></tr>     <tr><td>name: </td><td><? echo $row[name]; ?></td></tr>     </table>      <?  } ?>

i have tried many different ways achieve this, using joins on second select query, nothing seems work. know second query correct, because if hard code "$new_id = p12803 ;" second query grabs data.

any appreciated.

thanks

you need move second loop inside first loop otherwise $row[protein_id] return null. furthermore assign $row both fetching, changed second $rowa. change to

while ($row = mysql_fetch_assoc($rs_result)) {  ?>  <table class="table"> <tr><td>gene: </td><td><? echo $row[gene_id]; ?></td></tr> <tr><td>protein: </td><td><? echo $row[protein_id]; ?></td></tr> <tr><td>disease: </td><td><? echo $row[disease_id]; ?></td></tr> </table>  <?php      $new_id = $row[protein_id];      $sqla = "select * table_b     protein_asc='".$new_id."'";      $rsa_result = mysql_query ($sqla,$connect);       while ($rowa = mysql_fetch_assoc($rsa_result)) {      ?>      <table class="table">     <tr><td>synonym: </td><td><? echo $rowa[synonym]; ?></td></tr>     <tr><td>name: </td><td><? echo $rowa[name]; ?></td></tr>     </table>      <?  } ?>  <?  } ?>

then remember mysql_ functions deprecated advise switch mysqli or pdo , indeed @ risk of sql injection, have here how can prevent sql injection in php?. should use prepared statment avoid risk


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