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> Найдено сообщений: 2
gnerse Отправлено: 12 Августа, 2012 - 15:21:41 • Тема: Вывод из базы данных • Форум: Вопросы новичков

Ответов: 12
Просмотров: 513
Hello,
Please help me to solve this problem:

I use the following script for <search> option and it works fine.

//search

if (isset($_POST["skey"])) {

$sql_auth = '';
if ($_POST["sath"] != '' and $_POST["sath"] != 'all')
$sql_auth = "AuthorID IN (SELECT AuthorID FROM <Table Name> WHERE Author_Am LIKE '%$_POST[sath]%') AND";

if ($_POST["stit"] != '' and $_POST["stit"] != 'all')
$sql_tit = "Title LIKE '%$_POST[stit]%' AND";

$disp = '<br>';

$sql = "SELECT CatalogID, AuthorID, Title, PrintYear, WherePrintedID, PrintHouseID, Epilog, Memory FROM <table name> WHERE $sql_auth $sql_tit ( Contents LIKE '%$_POST[skey]%' OR Description LIKE '%$_POST[skey]%' OR Memory LIKE '%$_POST[skey]%' OR Epilog LIKE '%$_POST[skey]%' )";
$res = mysql_query($sql);

if (mysql_num_rows($res) == 0 ) {

$disp .= "No results for <b>$_POST[sath] $_POST[stit] $_POST[skey]</b>";
}
else {

$disp .= "<form name=f1 method=post>";
$disp .= "<input type=hidden name=c_id value=''>";
$disp .= "<input type=hidden name=c_ab_id value=''>";
$disp .= "<input type=hidden name=b value=''>";
$disp .= "Search results for <b>$_POST[skey]</b><br><br>\n";


while ($line = mysql_fetch_row($res)) {

if ($wr_pr = mysql_query("SELECT WherePrinted_Am FROM <table name> WHERE WherePrintedID = $line[4]")) {
$wrpr= mysql_fetch_row($wr_pr);
}

if ($pr_hs =mysql_query("SELECT PrintHouse_Am FROM <table name> WHERE PrintHouseID = $line[5]")) {
$prhs = mysql_fetch_row($pr_hs);
}

$disp .= "<b>TITLE:</b> $line[2]<br>\n";
if ($rs_a = mysql_query("SELECT Author_Am FROM <table name> WHERE AuthorID = $line[1]")) {
$athr = mysql_fetch_row($rs_a);
$disp .= "<b>AUTHOR:</b> $athr[0]<br>\n";
}

$disp .= "$wrpr[0], $prhs[0], $line[3]: <br><br>";
$disp .= "<button class=butt onClick=\" document.f1.b.value = '$_POST[skey]'; document.f1.c_id.value = $line[0]; document.f1.submit(); \">view</button> &nbsp; <button class=butt onClick=\" document.f1.c_ab_id.value = $line[0]; document.f1.submit(); \">add to basket</button> <hr>\n";
}
$disp .= "</form>";
}
}
================================ ============================
I wanted to use the above code for filtering results from another script:


// Proper Names View

if (isset($_POST["w_id"]) and $_POST["w_id"] != '') {

$disp .= "<br><table width=590>";

$disp .= "<form action=index.php name=f1 method=post>";
$disp .= "<input type=hidden name=skey value=''>";
$disp .= "<input type=hidden name=c_id value=''>";
$disp .= "<input type=hidden name=b value=''>";

$sql = "SELECT * FROM <Table Name> WHERE PrNmID = $_POST[w_id]";
$res = mysql_query($sql);
$line = mysql_fetch_array($res, MYSQL_ASSOC);

if ($line['GeneralName_Am'] != '') $disp .= "<tr><td><b>$tt_name</b></td><td><b>$line[GeneralName_Am]</b></td></tr><br>\n";
if ($line['Description'] != '') $disp .= "<tr><td><b>$tt_origin</b></td><td>$line[Description]</td></tr><br> \n";


$sql_exp = "SELECT PrNmExpID, Description, PrNmID FROM <Table Name> WHERE PrNmID =$line[PrNmID]";
$rs_exp = mysql_query($sql_exp);
$ln_exp = mysql_fetch_array($rs_exp, MYSQL_ASSOC);


// if ($ln['Description'] != '') $disp .= "<tr><td><b>$tt_apply</b></td><td>$ln[Description]</td></tr><br> \n";

$sql_abc = "SELECT CatalogID, Memory FROM <Table Name> WHERE Memory LIKE '%$line[GeneralName_Am]%'";



$rs_abc = mysql_query($sql_abc);
$ln_abc = mysql_fetch_array($rs_abc, MYSQL_ASSOC);


if ($ln_abc['Memory'] != '') $disp .= "<tr><td><b>$tt_apply</b></td><td>Reference";
$disp .= "<button class=butt onClick=\" document.f1.b.value = '$_POST[skey]'; document.f1.c_id.value = $ln_abc[CatalogID]; document.f1.submit(); \">view</button></td></tr><br><hr>\n";

}

================================ ============================

When I run the last I receice the whole database results. It seems to me that I didnt defined skey correctly, since the search script needs this parameter: Please help me to solve this problem. Thank you in advanced. George
gnerse Отправлено: 12 Августа, 2012 - 14:46:47 • Тема: Редакторы PHP • Форум: Программное обеспечение

Ответов: 421
Просмотров: 255124
Hello,
Please help me to solve this problem:

I use the following script for <search> option and it works fine.

//search

if (isset($_POST["skey"])) {

$sql_auth = '';
if ($_POST["sath"] != '' and $_POST["sath"] != 'all')
$sql_auth = "AuthorID IN (SELECT AuthorID FROM <Table Name> WHERE Author_Am LIKE '%$_POST[sath]%') AND";

if ($_POST["stit"] != '' and $_POST["stit"] != 'all')
$sql_tit = "Title LIKE '%$_POST[stit]%' AND";

$disp = '<br>';

$sql = "SELECT CatalogID, AuthorID, Title, PrintYear, WherePrintedID, PrintHouseID, Epilog, Memory FROM <table name> WHERE $sql_auth $sql_tit ( Contents LIKE '%$_POST[skey]%' OR Description LIKE '%$_POST[skey]%' OR Memory LIKE '%$_POST[skey]%' OR Epilog LIKE '%$_POST[skey]%' )";
$res = mysql_query($sql);

if (mysql_num_rows($res) == 0 ) {

$disp .= "No results for <b>$_POST[sath] $_POST[stit] $_POST[skey]</b>";
}
else {

$disp .= "<form name=f1 method=post>";
$disp .= "<input type=hidden name=c_id value=''>";
$disp .= "<input type=hidden name=c_ab_id value=''>";
$disp .= "<input type=hidden name=b value=''>";
$disp .= "Search results for <b>$_POST[skey]</b><br><br>\n";


while ($line = mysql_fetch_row($res)) {

if ($wr_pr = mysql_query("SELECT WherePrinted_Am FROM <table name> WHERE WherePrintedID = $line[4]")) {
$wrpr= mysql_fetch_row($wr_pr);
}

if ($pr_hs =mysql_query("SELECT PrintHouse_Am FROM <table name> WHERE PrintHouseID = $line[5]")) {
$prhs = mysql_fetch_row($pr_hs);
}

$disp .= "<b>TITLE:</b> $line[2]<br>\n";
if ($rs_a = mysql_query("SELECT Author_Am FROM <table name> WHERE AuthorID = $line[1]")) {
$athr = mysql_fetch_row($rs_a);
$disp .= "<b>AUTHOR:</b> $athr[0]<br>\n";
}

$disp .= "$wrpr[0], $prhs[0], $line[3]: <br><br>";
$disp .= "<button class=butt onClick=\" document.f1.b.value = '$_POST[skey]'; document.f1.c_id.value = $line[0]; document.f1.submit(); \">view</button> &nbsp; <button class=butt onClick=\" document.f1.c_ab_id.value = $line[0]; document.f1.submit(); \">add to basket</button> <hr>\n";
}
$disp .= "</form>";
}
}
================================ ============================
I wanted to use the above code for filtering results from another script:


// Proper Names View

if (isset($_POST["w_id"]) and $_POST["w_id"] != '') {

$disp .= "<br><table width=590>";

$disp .= "<form action=index.php name=f1 method=post>";
$disp .= "<input type=hidden name=skey value=''>";
$disp .= "<input type=hidden name=c_id value=''>";
$disp .= "<input type=hidden name=b value=''>";

$sql = "SELECT * FROM <Table Name> WHERE PrNmID = $_POST[w_id]";
$res = mysql_query($sql);
$line = mysql_fetch_array($res, MYSQL_ASSOC);

if ($line['GeneralName_Am'] != '') $disp .= "<tr><td><b>$tt_name</b></td><td><b>$line[GeneralName_Am]</b></td></tr><br>\n";
if ($line['Description'] != '') $disp .= "<tr><td><b>$tt_origin</b></td><td>$line[Description]</td></tr><br> \n";


$sql_exp = "SELECT PrNmExpID, Description, PrNmID FROM <Table Name> WHERE PrNmID =$line[PrNmID]";
$rs_exp = mysql_query($sql_exp);
$ln_exp = mysql_fetch_array($rs_exp, MYSQL_ASSOC);


// if ($ln['Description'] != '') $disp .= "<tr><td><b>$tt_apply</b></td><td>$ln[Description]</td></tr><br> \n";

$sql_abc = "SELECT CatalogID, Memory FROM <Table Name> WHERE Memory LIKE '%$line[GeneralName_Am]%'";



$rs_abc = mysql_query($sql_abc);
$ln_abc = mysql_fetch_array($rs_abc, MYSQL_ASSOC);


if ($ln_abc['Memory'] != '') $disp .= "<tr><td><b>$tt_apply</b></td><td>Reference";
$disp .= "<button class=butt onClick=\" document.f1.b.value = '$_POST[skey]'; document.f1.c_id.value = $ln_abc[CatalogID]; document.f1.submit(); \">view</button></td></tr><br><hr>\n";

}

================================ ============================

When I run the last I receice the whole database results. It seems to me that I didnt defined skey correctly, since the search script needs this parameter: Please help me to solve this problem. Thank you in advanced. George

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