blocks|key|333530|text|+$table++=+'your+table+name';
+$column+=+'q6'
+$add+=+mysql_query("ALTER+TABLE+$table+ADD+$column+VARCHAR(+255+)+NOT+NULL");|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|333531|您可以将VARCHAR(+255+)+NOT+NULL更改为任何您想要的datatype。|unstyled|offset|length|style|CODE|333532|entityMap^0|0|4|N|10|8|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code> $table = 'your table name';
 $column = 'q6'
 $add = mysql_query("ALTER TABLE $table ADD $column VARCHAR( 255 ) NOT NULL");
</code></pre>

you can change <code>VARCHAR( 255 ) NOT NULL</code> into what ever <code>datatype</code> you want.

blocks|key|333560|text|类似于：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|333561|$db+=+mysqli_connect("localhost",+"user",+"password",+"database");
$name+=+$db->mysqli_real_escape_string($name);
$query+=+'ALTER+TABLE+assesment+ADD+'+.+$name+.+'+TINYINT+NOT+NULL+DEFAULT+\'0\'';
if($db->query($query))+{
++++echo+"It+worked";
}|code-block|syntax|javascript|333562|还没有测试过，但应该可以工作。|333563|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Something like:

<pre><code>$db = mysqli_connect("localhost", "user", "password", "database");
$name = $db-&gt;mysqli_real_escape_string($name);
$query = 'ALTER TABLE assesment ADD ' . $name . ' TINYINT NOT NULL DEFAULT \'0\'';
if($db-&gt;query($query)) {
 echo "It worked";
}
</code></pre>

Haven't tested it but should work.

blocks|key|333614|text|根据您的注释，只有在：mysql_query("SELECT+*+FROM+assessment");返回false时，您才会添加新列。这可能不是你想要的。尝试删除“！”在第一个'if‘语句中的$sql前面。因此，您的代码将如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|333615|$sql=mysql_query("SELECT+*+FROM+assessment");
if+($sql)+{
+mysql_query("ALTER+TABLE+assessment+ADD+q6+INT(1)+NOT+NULL+AFTER+q5");
+echo+'Q6+created';+
}else...|code-block|syntax|javascript|333616|entityMap^0|B|14|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Based on your comment it looks like your'e only adding the new column if: <code>mysql_query("SELECT * FROM assessment");</code> returns false. That's probably not what you wanted. Try removing the '!' on front of $sql in the first 'if' statement. So your code will look like:

<pre><code>$sql=mysql_query("SELECT * FROM assessment");
if ($sql) {
 mysql_query("ALTER TABLE assessment ADD q6 INT(1) NOT NULL AFTER q5");
 echo 'Q6 created'; 
}else...
</code></pre>

blocks|key|333647|text|你的桌子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|333648|q1+%7C+q2+%7C+q3+%7C+q4+%7C+q5|code-block|syntax|javascript|333649|你也可以这样做|333650|ALTER+TABLE+yourtable+ADD+q6+VARCHAR(+255+)+after+q5|333651|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

your table:

<pre><code>q1 | q2 | q3 | q4 | q5
</code></pre>

you can also do 

<pre><code>ALTER TABLE yourtable ADD q6 VARCHAR( 255 ) after q5
</code></pre>

blocks|key|116192|text|对于WORDPRESS：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|116193|global+$wpdb;


$your_table++=+$wpdb->prefix.+'My_Table_Name';
$your_column+=++++++++++++++++'My_Column_Name';+

if+(!in_array($your_column,+$wpdb->get_col(+"DESC+"+.+$your_table,+0+)+)){++$result=+$wpdb->query(
++++"ALTER+++++TABLE+$your_table+++++ADD+$your_column+++++VARCHAR(100)+++++CHARACTER+SET+utf8+++++NOT+NULL+++++"++//you+can+add+positioning+phraze:+"AFTER+My_another_column"
);}|code-block|syntax|javascript|116194|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

for WORDPRESS:

<pre><code>global $wpdb;


$your_table = $wpdb-&gt;prefix. 'My_Table_Name';
$your_column = 'My_Column_Name'; 

if (!in_array($your_column, $wpdb-&gt;get_col( "DESC " . $your_table, 0 ) )){ $result= $wpdb-&gt;query(
 "ALTER TABLE $your_table ADD $your_column VARCHAR(100) CHARACTER SET utf8 NOT NULL " //you can add positioning phraze: "AFTER My_another_column"
);}
</code></pre>

blocks|key|116205|text|ALTER+TABLE+`stor`+ADD+`buy_price`+INT(20)+NOT+NULL+;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|116206|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>ALTER TABLE `stor` ADD `buy_price` INT(20) NOT NULL ;
</code></pre>

blocks|key|116292|text|PHP代码中ALTER+TABLE的问题在下面这一行中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|116293|mysql_query("ALTER+TABLE+assessment+newq+INT(1)+NOT+NULL+AFTER+q10")；|116294|它应该在q5之后，因为您的sample表中没有q10。因此，在q5之后，它变成了ALTER+TABLE+assessment+newq+INT(1)+NOT+NULL；|116295|在Skyvia中我自己的表中尝试了相同的逻辑，它应该可以如下所示。我在Title表中的To列之后添加了description列。|116296|Skyvia+-到https://skyvia.com/query/online-mysql-query-builder的链接|offset|length|116297|图像alter-table-add-column|116298|📷|atomic|116299|​|116300|entityMap|0|LINK|mutability|MUTABLE|url|https://skyvia.com/query/online-mysql-query-builder|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/c330491f16de50552ee61c2d15466b78.png|imageAlt^0|0|0|0|0|9|1F|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|18|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|19|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|1A|8|@]|9|@$J|1B|K|1C|1|1D]]|A|$]]|$1|L|3|M|5|6|7|1E|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1F|8|@]|9|@$J|1G|K|1H|1|1I]]|A|$]]|$1|Q|3|R|5|6|7|1J|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]|10|$5|11|W|12|A|$13|14|15|-4]]]]

The problem with the ALTER TABLE in the PHP code is in this line:
mysql_query(&quot;ALTER TABLE assessment ADD newq INT(1) NOT NULL AFTER q10&quot;);
It should be AFTER q5 since there is no q10 in your table sample. So, it becomes
ALTER TABLE assessment ADD newq INT(1) NOT NULL AFTER q5;
Tried the same logic in my own table in [Skyvia] and it should work as seen below.
I added the description column after the To column in the Title table.
[Skyvia] - the link to <a href="https://skyvia.com/query/online-mysql-query-builder" rel="nofollow noreferrer">https://skyvia.com/query/online-mysql-query-builder</a>
[image alter-table-add-column]<a href="https://i.stack.imgur.com/8p4We.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8p4We.png" alt="enter image description here" /></a>

I am trying to add a new column to my MYSQL table using PHP. I am unsure how to alter my table so that the new column is created. In my assessment table I have:
<pre><code>assessmentid | q1 | q2 | q3 | q4 | q5 
</code></pre>
Say I have a page with a textbox and I type <code>q6</code> in to the textbox and press a button then the table is updated to:
<pre><code>assessmentid | q1 | q2 | q3 | q4 | q5 | q6
</code></pre>
My code:
<pre><code>&lt;?php 
 mysql_query(&quot;ALTER TABLE `assessment` ADD newq INT(1) NOT NULL AFTER `q10`&quot;);
?&gt;
 &lt;form method=&quot;post&quot; action=&quot;&quot;&gt;
 &lt;input type=&quot;text&quot; name=&quot;newq&quot; size=&quot;20&quot;&gt;
 &lt;input type=&quot;submit&quot; name=&quot;submit&quot; value=&quot;Submit&quot;&gt;
</code></pre>

How to add new column to MYSQL table?

我正在尝试使用PHP向我的MYSQL表添加一个新列。我不确定如何更改我的表，以便创建新列。在我的评估表中，我有：assessmentid | q1 | q2 | q3 | q4 | q5 假设我有一个带有文本框的页面，我在文本框中键入q6并按下一个按钮，然后表格被更新为：assessmentid | q1 | q2 |...

问如何在MYSQL表中添加新列？
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问如何在MYSQL表中添加新列？
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