我有一个产品主表和包含产品属性的各种其他表,查询:
select p.description, category.value, colour.value, wood.value, brand.value, type.value, fabric.value, model.value from product_master p, category, colour, wood, brand, type, fabric, model where p.category_code=category.category_code and p.colour_code = colour.colour_code and p.wood_code = wood.wood_code and p.brand_code = brand.brand_code and p.type_code = type.type_code and p.fabric_code = fabric.fabric_code and p.model_code = model.model_code
在pgAdmin中正常工作但在php中它只提供了2列,我通过AJAX获得结果
我的PHP代码是
<?php
// Connecting, selecting database
$dbconn = pg_connect("host=***** dbname=*** user=*** password=***")
or die('Could not connect: ' . pg_last_error());
// Performing SQL query
$query = ' select p.description, category.value, colour.value, wood.value, brand.value, type.value, fabric.value, model.value from product_master p, category, colour, wood, brand, type, fabric, model where p.category_code=category.category_code and p.colour_code = colour.colour_code and p.wood_code = wood.wood_code and p.brand_code = brand.brand_code and p.type_code = type.type_code and p.fabric_code = fabric.fabric_code and p.model_code = model.model_code ';
$result = pg_query($query) or die('Query failed: ' . pg_last_error());
echo pg_affected_rows($result) ;
echo "\n";
echo pg_num_fields($result);
// Printing results in HTML
echo "<table>\n";
while ($line = pg_fetch_array($result, null, PGSQL_ASSOC)) {
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";
// Free resultset
pg_free_result($result);
// Closing connection
pg_close($dbconn);
?>
答案 0 :(得分:1)
您错误地使用了pg_fetch_array()。您不能将第二个参数作为NULL传递,因为它指示正在读取哪一行。
试试这个:
while ($line = pg_fetch_array($result)) {
echo "\t<tr>\n";
foreach ($line as $col_value)
echo "\t\t<td>$col_value</td>\n";
echo "\t</tr>\n";
}